MATH 4140/MATH 5140: Abstract Algebra II

Farid Aliniaeifard April 30, 2018

Contents

1 Rings and Isomorphism (See Chapter 3 of the textbook) 2 1.1 Deﬁnitions and examples ...... 2 1.2 Units and Zero-divisors ...... 6 1.3 Useful facts about ﬁelds and integral domains ...... 6 1.4 Homomorphism and isomorphism ...... 7

2 Polynomials Arithmetic and the Division algorithm 9

3 Ideals and Quotient Rings 11 3.1 Finitely Generated Ideals ...... 12 3.2 Congruence ...... 13 3.3 Quotient Ring ...... 13

4 The Structure of R/I When I Is Prime or Maximal 16

5 EPU 17 √ 6 Z[ d], an integral domain which is not a UFD 22 7 The ﬁeld of quotients of an integral domain 24

8 If R is a UDF, so is R[x] 26

9 Vector Spaces 28

10 Simple Extensions 30

11 Algebraic Extensions 33

12 Splitting Fields 35

13 Separability 38

14 Finite Fields 40

15 The Galois Group 43

1 16 The Fundamental Theorem of Galois Theory 45 16.1 Galois Extensions ...... 47

17 Solvability by Radicals 51 17.1 Solvable groups ...... 51

18 Roots of Unity 52

19 Representation Theory 53 19.1 G-modules and Group algebras ...... 54 19.2 Action of a group on a set yields a G-module ...... 54

20 Reducibility 57

21 Inner product space 58

22 Maschke’s Theorem 59

1 Rings and Isomorphism (See Chapter 3 of the textbook)

1.1 Deﬁnitions and examples Deﬁnition. A ring is a nonempty set R equipped with two operations (usually written as addition (+) and multiplication(.) and we denote the ring with its operations by (R,+,.)) that satisfy the following axioms. For all a, b, c ∈ R:

(1) If a ∈ R and b ∈ R, then a + b ∈ R. [Closure for addition]

(2) a + (b + c) = (a + b) + c. [Associative addition]

(3) a + b = b + a. [Commutative addition]

(4) There is an element 0R in R such that a + 0R = a = 0R + a for every a ∈ R. [Zero element]

(5) For every a ∈ R, there exists an element b ∈ R such that a+b = 0 = b+a. [additive Inverse element]

(6) If a ∈ R and b ∈ R, then a.b ∈ R. [Closed for multiplication]

(7) a.(b.c) = (a.b).c

(8) a.(b + c) = a.b + a.c and (a + b).c = a.c + b.c [Distributive laws]

Remark. Note that axioms 1, 2, 3, 4, and 5 shows that (R, +) is an abelian group.

Example 1.1. Z, Zn, and M2(R) are rings. Deﬁnition. A commutative ring is a ring R that satisﬁes the axiom:

(9) a.b = b.a for all a, b ∈ R. [Commutative multiplication]

2 Deﬁnition. A ring with identity is a ring R that contains an element 1R satisfying this axiom:

(10) a.1R = a = 1R.a for all a ∈ R [Multiplicative identity]

Deﬁnition. An integral domain is a commutative ring R with identity 1R 6= 0 that satisﬁes this axiom:

(11) whenever a, b ∈ R and a.b = 0, then a = 0R or b = 0R.

The end of the lecture 1

3 Lecture 2, January 19, 2018

A ring is a nonempty set R together with two operations (+) and (.) such that

(i) (R,+) is an abelian group.

(ii) for all a, b ∈ R, a.b ∈ R.

(iii) for all a, b, c ∈ R, a.(b.c) = (a.b).c.

(iv) for all a, b, c ∈ R, we have a.(b + c) = a.b + a.c and (a + b).c = a.c + b.c.

Then a ring is commutative if a.b = b.a for all a, b ∈ R. It has an identity if there is an element 1 ∈ R such that a.1 = 1.a = a. An integral domain is a commutative ring with identity 1 that satisfy the axiom: if ab = 0, then a = 0 or b = 0.

Example 1.2. Z, R, and Q are integral domains. Z6 is a ring but not an integral domain. The elements of Z6 are {0, 1, 2, 3, 4, 5}. In Z6, we have 2 + 3 = 5, 5 + 4 = 3 and 5 + 3 = 2. Also, if we compute 2.3 = 0

we see that Z6 is not an integral domain.

Example 1.3. Let R be the ring of real numbers. Then R[x] is a ring with the following addition and multiplication: if

2 f(x) = 1 + 3x + x ∈ R[x] and 3 g(x) = −3x + x ∈ R[x], then f(x) + g(x) = 1 + x2 + x3 and

f(x)g(x) = −3x + x3 − 9x2 + 3x4 − 3x3 + x5 = −3x − 9x2 − 2x3 + 3x4 + x5.

Also, the polynomial e(x) = 1 is the identity element and 0(x) = 0 is the zero element.

Theorem 1.4. Let R and S be rings. Deﬁne the following addition and multiplication on the Cartisian Product R × S by

(r, s) + (r0 , s0 ) = (r + r0 , s + s0 ) and (r, s).(r0 , s0 ) = (r.r0 , s.s0 ).

Then R × S is a ring. If R and S are both commutative, so does R × S. If both R × S have identity then so does R × S.

In the following proposition we will present some properties of the elements of a ring.

Proposition 1.5. Let R be a ring and a, b ∈ R. Then

(i) 0.a = a.0 = 0.

4 (ii) (−a).b = a.(−b) = −(a.b).

(iii) if R has identity 1, then the identity is unique and (−a) = (−1).a.

Proof. 1. (i)

0.a = (0 + 0).a = 0.a + 0.a ⇒ 0.a = 0a + 0a ⇒ 0.a = 0.

2. (ii)

a.b + (−a).b = (a + (−a)).b = 0.b = 0 ⇒ a.b + (−a).b = 0 ⇒ (−a).b = −(ab).

The rest left to the reader.

Deﬁnition. A subset S of a ring R is a subring of R if it is a ring with the same addition and multiplication as R. To show that a subset S of R is a subring of R, you only need to check that S is nonempty and

(i) S is closed under multiplication.

(ii) S is closed under subtraction, i.e., if a, b ∈ R, then a − b ∈ R. √ √ Example 1.6. Deﬁne Z[ 2] = {a + b 2 : a, b ∈ Z} is a subring of R. √ √ √ √ Proof. Let a + b 2 ∈ Z[ 2] and c + d 2 ∈ Z[ 2]. Then √ √ √ √ (a + b 2)(c + d 2) = (ac + 2bd) + (ad + cb) 2 ∈ Z[ 2]

and √ √ √ √ (a + b 2) − (c + d 2) = (a − c) + (b − d) 2 ∈ Z[ 2]. √ Therefore Z[ 2] is a subring of R.

Deﬁnition. A Field is a commutative ring R with identity 1R 6= 0 that satisﬁes this axiom:

(12) For each a 6= 0R in R, there is an element b ∈ R such that ab = 1 = ba. The element b is called the inverse of a and is denoted by a−1.

Example 1.7. (i) Q, R and C are ﬁelds.

(ii) Zp when p is a prime number is a ﬁeld.

5 Week 2, Lecture 1

1.2 Units and Zero-divisors Deﬁnition. An element a in a ring R with identity is called a unit if there exists u ∈ R such that au = 1R = ua.

Example 1.8. (i) Units of Z are 1 and −1.

(ii) Units of Z6 are 1 and 5. In general the set of units of Zn is {k : k and n are coprime }.

(iii) Units of the rings of 2 × 2 matrices M2(R) are the invertible matrices. Deﬁnition. An element a in R is a zero-divisor provided that

(i) a 6= 0R.

(ii) There exists a nonzero element c ∈ R such that ac = 0R or ca = 0R. Remark. An integral domain contains no zero-divisors.

Example 1.9. (i) 2 and 3 are zero-divisors in Z6 since 2.3 = 0.

(ii) zero-divisors of the rings of 2 × 2 matrices M2(R) are the non-invertible matrices. a b Because if is a non-invertible matrix we have det(A) = ad − bc = 0, and c d a b d −b 0 0 we multiply = c d −c a 0 0

A ring D with identity 1D in which every non-zero element is a unit is called a Division ring. Note that any ﬁeld is a commutative division ring.

1.3 Useful facts about ﬁelds and integral domains

Theorem 1.10. Every finite field F is an integral domain. Proof. Every field is a commutative ring with identity, so it is enough to show that if ab = 0, then a = 0 or b = 0. Assume that ab = 0 but a 6= 0 and b 6= 0. Then

ab = 0 ⇒ a−1(ab) = 0 ⇒ 1.b = 0 ⇒ b = 0,

which is a contradiction.

Theorem 1.11. Every ﬁnite integral domain R is a ﬁeld.

Proof. As every integral domain is a commutative ring with identity, it is enough to show that every non-zero element in R has an inverse in R. Let a1, . . . , an be all of nonzero distinct elements of R. Let a be an arbitrary nonzero element in R. Then for any ai we have aai 6= 0 otherwise since R is in integral domain we have a = 0 or ai = 0 which is not possible. Therefore, aa1, . . . , aan are nonzero elements in R. Moreover, they are distinct because if for some distinct i and j, aai = aaj, then

aai − aaj = 0 ⇒ a(ai − aj) = 0.

6 Since R is an integral domain and a 6= 0, we must have ai −aj = 0, and so ai = aj, which is not possible. Therefore, for all distinct i and j we have aai 6= aaj. So aa1, . . . , aan are all of nonzero distinct elements of R. We can conclude that {a1, . . . , an} = {aa1, . . . , aan}. Since 1 ∈ {a1, . . . , an} = {aa1, . . . , aan}. Thus for some i we have aai = 1, and so −1 a = ai, i.e., a is invertible. It follows that every element of R has an inverse and so R is a ﬁeld.

1.4 Homomorphism and isomorphism Deﬁnition. Let R and S be rings. A function f : R → S is said to be a homomorphism if f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b) for all a, b ∈ R.

Deﬁnition. A ring R is isomorphic to a ring S (in symbols, R ∼= S) if there is a homomorphism f : R → S such that

(i) f is injective;

(ii) f is surjective.

7 Week 2, Lecture 2

The kernel of a homomorphism of rings f : R → S is the set

kerf = {r ∈ R : f(r) = 0}.

The image of f denoted by Imgf, is {s ∈ S : s = f(r) for some r ∈ R }. Theorem 1.12. Let f : R → S be a homomorphism of rings. Then

1. f(0R) = f(0S). 2. f(−a) = −f(a).

3. f(a − b) = f(a) − f(b).

If R is a ring with identity and f is surjective, then

4. S is a ring with identity f(1R). 5. Whenever u is a unit in R, then f(u) is a unit in S and f(u)−1 = f(u−1). Proof. Refer to the book for (1), (2), and (3). (4) We want to show that for any s ∈ S, we have sf(1R) = f(1R)s = s. Since f is surjective, there is an element r ∈ R such that f(r) = s. Therefore,

sf(1R) = f(r)f(1R) = f(r1R) = f(r) = s.

Similarly we have f(1R)s = s. (5) Since u is a unit in R, there is an element v ∈ R such that uv = vu = 1R. Therefore, f(uv) = f(1R). So, f(u)f(v) = f(v)f(u) = f(1R). Note that f(1R) is the identity of S, therefore, f(u) is a unit in S. Now we want to show that f(u)−1 = f(u−1). Note that −1 −1 −1 −1 uu = 1R, therefore, f(u)f(u ) = f(1R), which means that f(u) = f(u ). Example 1.13. Let a b R = : a, b ∈ . −b a R Show that R is a ring and is isomorphic to C the ring of complex numbers. Proof. Deﬁne a function as follows

f : R → C a b 7→ a + ib −b a

Clearly this function is well-deﬁned, surjective and one to one, so we only show that it is a ring homomorphism.

a b c d a + c b + d f( + ) = f( ) = (a+c)+i(c+d) = (a+ib)+(c+id) −b a −d c −(b + d) a + c

a b c d = f( ) + f( ) −b a −d c

8 a b c d ac − bd ad + bc f( ) = f( ) = (ac−bd)+(ad+bc)i = (a+ib)(c+id) −b a −d c −ad − bc ac − bd

a b c d = f( )f( ) −b a −d c

Corollary 1.14. If f : R → S is a homomorphism, then the image of f is a subring of S.

0 Proof. Note that f(0R) = 0S, so Imgf is nonempty. Let s, s ∈ Imgf. Then there are elements r, r0 ∈ R such that f(s) = r and f(r0 ) = s0 . Now,

ss0 = f(r)f(r0 ) = f(rr0 ) ∈ Imgf and s − s0 = f(r) − f(r0 ) = f(r − r0 ) ∈ Imgf. We conclude that Imgf is a subring of S.

2 Polynomials Arithmetic and the Division algorithm

Let R be any ring. A polynomial with coeﬃcients in R is an expression of the form

2 n a0 + a1x + a2x + ··· + anx ,

where n is a nonnegative integer and ai ∈ R. Let R[x] be the set of all polynomials in R[x]. Actually, R[x] is a subring of another ring T (we do not discuss the structure of T in this course). The element x sometimes called an indeterminate.

n Deﬁnition. Let f(x) = a0 + a1x + ... + anx ∈ R[x] such that an 6= 0. Then an is called the leading coeﬃcient of f(x). The degree of f(x) is the integer n, and we write degf(x) = n. We can consider elements of R as polynomials in R[x], and they are called constant polynomials. The polynomials of degree 0 in R[x] are precisely the constant polynomials. Note that 0R does not have a degree. Theorem 2.1. If R is an integral domain and 0 6= f, g ∈ R[x], then

degf(x)g(x) = degf(x) + degg(x).

n m Proof. Suppose that f(x) = a0 +a1x+...+anx (an 6= 0) and g(x) = b0 +b1x+...+bmx (am 6= 0). So degf(x) = n and degg(x) = m. Then

2 n+m f(x)g(x) = a0b0 + (a0b1 + a1b0)x + (a2b0 + a1b1 + a0b2)x + ... + anbmx .

Since R is an integral domain and an, bm 6= 0, we have that anbm 6= 0, and so degf(x)g(x) = n + m = degf(x) + degg(x).

Corollary 2.2. If R is an integral domain, so is R[x].

9 Proof. We have that R[x] is a commutative ring with identity. We will show that it does not have any nonzero zero-divisors. Let 0 6= f(x), g(x) ∈ R[x]. Let f(x) = a0 + a1x + n m ... + anx (an 6= 0) and g(x) = b0 + b1x + ... + bmx (am 6= 0). Then

2 n+m f(x)g(x) = a0b0 + (a0b1 + a1b0)x + (a2b0 + a1b1 + a0b2)x + ... + anbmx .

If f(x)g(x) = 0, then anbm = 0, which is impossible since R is an integral domain. Therefore, R[x] is an integral domain.

Corollary 2.3. Let R be a ring. If f(x), g(x) and f(x)g(x) are non-zero in R[x], then degf(x)g(x) ≤ degf(x)degg(x).

10 Week 2, Lecture 3

Corollary 2.4. Let R be an integral domain and f(x) ∈ R[x]. Then f(x) is a unit in R[x] if and only if f(x) is a constant polynomial that is a unit in R. In particular, F is a ﬁeld, then units in F[x] are the nonzero constant in F. Proof. Assume that f(x) is a unit in R[x], then there is a polynomial g(x) such that f(x)g(x) = 1. Since R is an integral domain, we have degf(x)+degg(x) = degf(x)g(x) = deg 1 = 0. Therefore, we must have both f(x) and g(x) are of degree 0, so they are constant, and actually they are units in R. The theorem above is not true if R is not an integral domain, for example, 5x + 1 ∈ Z25[x] is not a constant, however, it is a unit since (5x + 1)(20x + 1) = 1.

Theorem 2.5. Let F be a ﬁeld and f(x), g(x) ∈ F[x] with g(x) 6= 0. Then there exist unique polynomials q(x) and r(x) such that

f(x) = g(x)q(x) + r(x) and either r(x) = 0 or degr(x) < degg(x).

3 Ideals and Quotient Rings

An ideal of a ring R is a non-empty subset of R such that for all a, b ∈ I and r ∈ R,

1. a − b ∈ I

2. ra ∈ I and ar ∈ I

Example 3.1. Let T = {f : R → R : f is a function}. Deﬁne the following addition and multiplication for T . For all f, g ∈ T and a ∈ R,

(f + g)(a) = f(a) + g(a) fg(a) = f(a)g(a).

Then T is a ring with the above addition and multiplication. Show that the following set is an ideal of T ,

I = {f : R → R : f is a funtion and f(2) = 0}.

Proof. The set I is nonempty since 0 : R → R deﬁned by 0(r) = 0 for every r ∈ R, is in I. Moreover, for all f, g ∈ I and h ∈ I, we have that

(f − g)(2) = f(2) − g(2) = 0 − 0 = 0 and hf(2) = h(2)f(2) = h(2)0 = 0 and fh(2) = f(2)h(2) = 0h(2) = 0. Therefore, f − g, fh, hf ∈ I and so I is an ideal.

Deﬁnition. A nonempty subset of a ring R is a left (right) ideal if for all a, b ∈ I, and r ∈ R, a − b ∈ I and ra ∈ I(ar ∈ I).

Remark. Any ideal of R is a subring of R. Also, a left (right) ideal is a right (left) ideal in a commutative ring.

11 3.1 Finitely Generated Ideals Theorem 3.2. Let R be a commutative ring with identity and let c ∈ R. Deﬁne

I = {rc : r ∈ R}.

Then I is an ideal of R.

Proof. Note that 0 = 0c ∈ I, so I is nonempty. Moreover, for all r1c, r2c ∈ I and r ∈ R. Then r1c − r2c = (r1 − r2)c ∈ I and r(r1c) = (rr1)c ∈ I.

As R is commutative (r1c)r ∈ I. Therefore, I is an ideal of the commutative ring R. Deﬁnition. The ideal I deﬁned in the above theorem is called the principal ideal generated by c and is denoted by hci.

Example 3.3. Let

I = {All polynomials in Z[x] with even constant term }.

Then I is an ideal of Z[x] but I is not principal. Proof. First note that I is an ideal (show it). We now show that I is not principal, i.e., there is not any polynomial p(x) ∈ Z[x] such that

I = hp(x)i = {f(x)p(x): f(x) ∈ Z[x]}. On the contrary assume that I = hp(x)i. Then since 2 ∈ I, there is an polynomial f(x) ∈ Z[x] such that f(x)p(x) = 2. Since Z is an integral domain, we have degf(x) + degp(x) = degf(x)g(x) = deg2 = 0.

We can say that degp(x) = 0 and let p(x) = c. Since c|2, we have c = 2 or −2. Now, x ∈ I. Therefore, f(x)c = x for some f(x) ∈ Z[x]. By comparing the degrees, degf(x) = 1. So f(x) = a + bx for some a, b ∈ Z. Therefore, (a + bx)c = x, and so ac + bcx = x, which means that bc = 1. Thus c is invertible, which is a contradiction.

12 Week 3, Lecture 1

Theorem 3.4. Let R be a commutative ring with identity, and c1, . . . , cn ∈ R. Then the set I = {r1c1 + r2c2 + ... + rncn : r1, r2, . . . , rn ∈ R} is an ideal.

Definition. The ideal in the above theorem is called the ideal generated by c1, . . . , cn and denoted by hc1, ··· , cni. Definition. If an ideal can be generated by a finite number of elements in R, then I is a finitely generated ideal.

3.2 Congruence Deﬁnition. Let I be an ideal in R and a, b ∈ R. Then a is congruent to b modulo I [written a ≡ b(mod I)] if a − b ∈ I.

Example 3.5. Let f(x) = x2+6 and g(x) = 5x in I = {f : R → R : f is a funtion and f(2) = 0}. Then f(x) ≡ g(x)(mod I) because f(x) − g(x) = (x2 + 6) − 5x, (f − g)(2) = 0, and so f(x) − g(x) ∈ I.

3.3 Quotient Ring • Fix an ideal I of R. The relation ≡ (mod I) is reﬂexive, symmetric, and transitive. Therefore, it is an equivalence relation.

• The equivalence class containing a ∈ R, denoted by a + I, is the set

a + I = {b ∈ R : b ≡ a(mod I)}.

• The set a + I is called a (left) coset of I in R.

Theorem 3.6. Let R be a ring and I be an ideal of R. Then

1. a + I = {a + i : i ∈ I}.

2. Two cosets are either identical or disjoint, i.e., for two cosets a + I and b + I we have either a + I = b + I or a + I ∩ b + I = ∅.

Proof. 1. We have

a + I = {b ∈ R : b ≡ a(mod I)} = {b ∈ R : b − a ∈ I}

= {b ∈ R : b − a = i for some i ∈ I} = {b ∈ R : b = a + i for some i ∈ I} = {a + i : i ∈ I}.

13 2. Let a + I and b + I be two cosets of I in R. Assume that a + I ∩ b + I 6= ∅ and c ∈ a + I ∩ b + I. Then c = a + i for some i ∈ I and c = b + j for some j ∈ I. So

a + i = b + j ⇒ a − b = j − i ∈ I.

Let B ∈ b + I. Then B = b + k for some k ∈ I. Also,

a − (b + k) = j − i − k ∈ I,

and so B = b + k ∈ a + I. Therefore, a + I ⊆ b + I. Similarly, we can show that b + I ⊆ a + I, and so a + I = b + I.

Deﬁne R/I := {a + I : a ∈ R}. Theorem 3.7. 1. The set R/I is a ring with the following addition and multiplication.

+ : R/I × R/I → R/I (a + I, b + I) 7→ (a + b) + I

. : R/I × R/I → R/I (a + I, b + I) 7→ (ab) + I

Also, 0R/I = I. The ring R/I is called quotient ring or factor ring of R by I. 2. If R is commutative, then so is R/I.

3. If R has identity, then so is R/I. Proof. 1. First we should show that the functions + and . are well-deﬁned, and then other axioms are easy to check. Let a + I, b + I, c + I, d + I ∈ R/I. Assume that (a + I, b + I) = (c + I, d + I). Then a + I = c + I and b + I = d + I. Therefore, a − c, b − d ∈ I. Now,

(a + b) − (c + d) = (a − c) + (b − d) ∈ I.

Therefore, (a + b) + I = (c + d) + I. Also, ab − cd = ab − cb + cb − cd = (a − c)b + c(b − d) ∈ I. Therefore, (ab) + I = (cd) + I. We conclude that both + and . are well-deﬁned.

2. Note that (a + I)(b + I) = (ab) + I = (ba) + I = (b + I)(a + I).

3. If R has an identity 1R, we have 1R + I is the identity of R/I since

(1R + I)(a + I) = (1Ra) + I = a + I = (a1R) + I = (a + I)(1R + I).

14 Week 3, Lecture 2. Theorem 3.8. Let R be a ring. 1. Let S be a ring. If f : R → S is a homomorphism, then kerf is an ideal of R.

2. Every ideal of R is a kernel of a homomorphism f : R → S. Proof. 1. We want to show that kerf is an ideal. Note that kerf is not empty since f(0) = 0, so 0 ∈ kerf. Let t, s ∈ kerf and r ∈ R. Then

f(s − t) = f(s) − f(t) = 0 − 0 = 0,

so s − t ∈ kerf. Also,

f(rs) = f(r)f(s) = f(r)0 = 0 and f(sr) = f(s)f(r) = 0f(r) = 0,

so rs, sr ∈ kerf.

2. Let I be an ideal then there is a homomorphism π : R → R/I deﬁned by π(r) = r + I. Then

kerπ = {r ∈ R : r + I = I} = {r ∈ R : r ∈ I} = I.

Theorem 3.9. Let f : R → S be a homomorphism. Then kerf = {0} if and only if f is injective. Proof. First assume that ker f = {0}. Then if f(a) = f(b), we have f(a) − f(b) = 0, and so f(a − b) = 0. Since kerf = 0, it follows that a − b = 0, i.e., a = b. Conversely, let a ∈ ker f, then we have f(a) = 0 = f(0). As f is injective, we must have a = 0. Theorem 3.10 (First Isomorphism Theorem). Let f : R → S be a homomorphism. Then R/kerf ∼= Imgf. Proof. Deﬁne f : R/I → Img f such that f(r + I) = f(r). We must show that f is a isomorphism, i.e, it is a one-to-one and surjective homomorphism. First we show that f is well-deﬁned. Let a+ker f = b+ker f, then a−b ∈ ker f, and so f(a−b) = 0. Therefore, f(a) − f(b) = 0, and this implies that f(a) = f(b). It is clear that f is surjective since Img f = Img f. To show that f is one-to-one, note that if f(a+ker f) = 0, then f(a) = 0, that is a ∈ ker f and so a + ker f = ker f = 0R/ker f . Therefore, by the previous theorem we have that f is one-to-one. Moreover,

f((a + ker f)(b + ker f)) = f(ab + ker f) = f(ab) = f(a)f(b) = f(a + ker f)f(b + ker f).

Similarly, we can show that f((a + ker f) + (b + ker f)) = f(a + ker f) + f(b + ker f). We can now conclude that f is an isomorphism. Theorem 3.11 (The Second Isomorphism Theorem). Let I and J be ideals in a ring R. Then I I + J ∼= . I ∩ J J 15 Theorem 3.12 (The Third Isomorphism Theorem). Let I and K be ideals of R I such that K ⊆ I. Then K is an ideal of R/K, also R/K ∼= R/I. I/K

Theorem 3.13 (The Forth Isomorphism Theorem). If f : R → S is a surjective homomorphism of rings with kernel K, then there is a bijection from the set of all ideals of S to the set of all ideals of R that contains K. (Now ask yourself what are the ideals of R/I?)

4 The Structure of R/I When I Is Prime or Maximal

An ideal P in a commutative ring R is said to be prime if P 6= R and whenever bc ∈ P , then b ∈ P or c ∈ P .

Theorem 4.1. Let P be an ideal in a commutative ring with identity. Then P is a prime ideal if and only if the quotient ring R/P is an integral domain.

Proof. Let P be a prime ideal. Note that 1R/P 6= 0, otherwise 1R + P = 0 + P , which implies that 1R ∈ P , and so R = P , a contradiction.Now we show that R does not have any zero-divisors. Assume on the contrary that a + P and b + P are both non-zero, but (a + P )(b + P ) = 0, then ab + P = P , and so ab ∈ P . It follows that a ∈ P or b ∈ P , which means a ∈ P or b ∈ P . Conversely, if R/P is an integral domain, then 1R + P 6= P , and so P 6= R. Assume that ab ∈ P but a 6∈ P and b 6∈ P , then (a + P )(b + P ) = (ab) + P = P = 0R/P , which means R/P has a zero-divisor, a contradiction.

Deﬁnition. An ideal M in a ring R is said to be maximal if M 6= R and whenever J is an ideal such that M ⊆ J ⊆ R, then either M = J or J = R.

Theorem 4.2. Let M be an ideal in a commutative ring with identity. Then M is a maximal ideal if and only if the quotient ring R/M is ﬁeld.

Proof. Let M be a maximal ideal. With the same argument as in the above theorem, we have that 1R/M 6= 0R/M . Now we show that every non-zero element in R/M is a unit element. Let a + M ∈ R/M. We first show that the ideal ha + Mi = R/M. On the contrary assume that ha+Mi= 6 R/M, By the forth isomorphism theorem we have ha+Mi is equal to some ideal J/M of R/M. Note that J is an ideal that containing M. Since M is maximal we have to have J = R, which is equivalent to say that ha + Mi = R/M, and so there is an element b + M such that (b + M)(a + M) = 1 + M. Therefore, every element in R/M has an inverse and so R/M is a field. Conversely, assume that R/M is a field. Since 1R/M 6= 0 + M we have that M 6= R. Now assume that there is an ideal J such that M ⊂ J. Then J/M is an ideal of R/M (again by forth isomorphism theorem), but we have R/M is a field and every element is invertible so there is a nonzero and so an invertible element in a + M ∈ J/M. Therefore, J/M = R/M and so R = J. Therefore, M is a maximal ideal.

Corollary 4.3. If M is a maximal ideal of R, then M is prime too.

16 5 EPU

Deﬁnition. Let R be a commutative ring with identity. Let a and b be in R.

• An element a ∈ R divides b ∈ R, written a|b, if b = ac for some c ∈ R.

• Two elements a and b are said to be associate if a|b and b|a.

• A nonzero nonunit element p ∈ R is said to be prime if p|ab, then p|a or p|b.

• A nonzero nonunit element p ∈ R is said to be irreducible if a = rs, then r or s is a unit.

Theorem 5.1. Let a, b and u be elements of a commutative ring R with identity.

1. a|b if and only if hbi ⊆ hai.

2. a and b are associate if and only if hai = hbi.

3. u is a unit if and only if u|r for all r ∈ R.

4. u is a unit if and only if hui = R.

5. The relation ”a is an associate of b” is an equivalence relation on R.

6. If a = br with r ∈ R a unit, then a and b are associates. If R is an integral domain, the converse is true.

Note that 2 is a prime element in Z6 but it is not an irreducible element since 2 = 2.4. Deﬁnition. An integral domain R is a unique factorization domain (UFO) provided that every nonzero, nonunit element of R is the product of irreducible elements, and this factorization is unique up to associates; that is, if

p1p2 . . . pr = q1q2 . . . qs with each pi and qj irreducible, then r = s and, after reordering and relabeling if necessary,

pi is an associate of qi for i = 1, 2, . . . , r.

Deﬁnition. A principal ideal ring is a ring in which every ideal is a principal ideal. A principal ideal ring which is also an integral domain is called a principal ideal domain.

Deﬁnition. An integral domain R is a Euclidean domain if there is a function δ from the nonzero elements of R to the nonnegative integers with these properties:

• If a and b are nonzero elements of R, then δ(a) ≤ δ(ab).

• If a, b ∈ R and b 6= 0, then there exist q, r ∈ R such that a = bq +r and either r = 0 or δ(r) < δ(b).

Example 5.2. 1. Every ﬁeld is an integral domain with the function δ given by δ(x) = 1 for all x in the ﬁeld.

2. Z is a Euclidean domain with the function δ given by δ(a) = |a|.

17 3. F[x] the polynomials with coeﬃcients in the ﬁeld F is a Euclidean domain with the function δ given by δ(f(x)) = deg(f(x)).

4. The ring of Gaussian integers

Z[i] = {s + ti : s, t ∈ Z} is a Euclidean domain with the function δ given by δ(s + ti) = s2 + t2.

18 Week 4, Lecture 2.

Theorem 5.3. Let p and c be non-zero elements in an integral domain R.

1. p is prime if and only if hpi is nonzero prime.

2. c is irreducible if and only if hci is maximal in the set S of all proper principal ideals of R.

3. Every prime element of R is irreducible.

4. If R is a principal ideal domain, then p is prime if and only if p is irreducible.

5. Every associate of an irreducible (resp. prime) element of R is irreducible (resp. prime).

6. The only divisors of an irreducible element of R are its associates and the units of R.

Proof. 1. Let p be a prime element and ab ∈ hpi. Then for some r ∈ R, we have pr = ab, i.e., p|ab and since p is a prime element, it follows that p|a or p|b, that is a ∈ hpi or b ∈ hpi. Conversely, assume that hpi is a prime ideal and p|ab, then ab ∈ hpi, and since hpi is a prime ideal, we have a or b is in hpi. Therefore, p|a or p|b.

2. Assume that c is an irreducible elements and there is a proper principle ideal hdi that contains hci, then d|c, and so da = c for some a ∈ R, since c is an irreducible element we must have a is a unit and so hci = hdi. Conversely, assume that hci is maximal in the set S of all proper principal ideals of R. Then if c = rs and none of r and s are units, we have hci = hri. Then there is a unit u such that c = ru. Therefore, ru = rs and so u = s.

3. If p is a prime element and p = rs, then p|r or p|s. Without loss of generality assume that p|s, then since s|p, we have hpi = hsi, and since they are elements of integral domain, same as previous part, we have r is a unit.

4. By (3) every prime is irreducible in an integral domain, so we only need to show that every irreducible is prime in a principal ideal domain. Let p be an irreducible element, then by (2) hpi is a maximal ideal in R, and so it is a prime ideal. Therefore, by (1) p is a prime element.

5. Let p be an irreducible element. Then if q is associate to p by previous theorem part (6), there is a unit u such that p = qu. Assume that q is not an irreducible element, then q = ab for some nonzero nonunit elements a and b, and so p = (ua)b. Note that ua is not unit because otherwise a is a unit. Therefore, q is not an irreducible element, a contradiction.

6. If r|p where p is an irreducible element, then we have rs = p for some s ∈ R. Since p is irreducible, r or s is a unit, which means either r is a unit or r is an associate of p.

19 Week 4, Lecture 3 Example 5.4. An example of a ring that is an integral domain and it has some irreducible elements that√ are not prime √ √ Let R be the subring {a+b 10 : a, b ∈ Z} of real numbers. Note that 2, 3, 4+ 10, 4− 10 are irreducible elements but not prime elements. Also, note that this subring also is not a UFD. Moreover, in this subring every element can be factored into irreducible elements but it is not unique. Proposition 5.5. Every irreducible element in a UFD is a prime element. Proof. Let p be an irreducible element in a UFD, say R. If ab ∈ hpi for some a, b ∈ R, then pr = ab. Now factor r = r1 . . . rk, a = a1 . . . at, and b = b1 . . . bt into irreducible elements. Then p must be equal to some ai or bj which means p|a or p|b, and so a ∈ hpi or b ∈ hpi. Therefore, hpi is a prime ideal and so p is a prime element. Our goal now is to show that every PID is a UFD. In order to prove that we need the following lemma. Lemma 5.6. If R is a principal ideal domain and

ha1i ⊆ ha2i ⊆ ... is a chain of ideals in R, then for some positive integer n, haji = hani for all j ≥ n. Proof. Let [ I = haii. i=1 Then note that I is an ideal of R. Since R is a principal ideal domain, there is an element c ∈ R such that I = hci. Since c ∈ I, there is an hani such that c ∈ hani. Therefore, we must have hani = hci.

Theorem 5.7. Every PID is a UFD. Proof. Assume that in the PID R there is an element a that can not be written as the product of irreducible elements. So we can write a = a1b1 such that a1 and b1 are not units and at least one of a1 or b1 can not be written as product of irreducible elements. WLOG assume that we can not write a1 as a product of prime elements. Note that hai ha1i, because otherwise b1 is a unit. Now we repeat the process for a1. So, we can write a1 = a2b2 such that ha1i ha2i. If we continue this process we will have the following chain that never end

ha1i ha2i ha3i ··· that is a contradiction since the previous theorem stated that any chain of ideal in a PID is stable. Now we will show that this factorization is unique up to associates and reordering. Let a be an arbitrary element and we write a as a product of irreducible elements in the following ways: p1p2 ··· ps = a = q1q2 ··· qr.

20 Without loss of generality assume that r ≥ s. Note that p1|q1q2 ··· qr. Therefore, Since in a PID every irreducible is a prime, we have that p1|qi for some i. After rearranging and relabeling the qi’s if necessary, we may assume that p1|q1. Therefore, p1u1 = q1 and as q1 is an irreducible element we must have u1 is a unit. Therefore, q1 and p1 are associate. So we can write p1p2 ··· ps = a = u1p1q2 ··· qr, for some unit u1. By the cancellation law we have

p2 ··· ps = (u1q2) ··· qr.

Note that u1q2 also is an irreducible element. Continue the same argument until there is no more pi’s left. So if s 6= r, we have 1 = u1u2 . . . usqs+1 . . . qr−1qr, and so qr is a unit, a contradiction. Therefore, we must have r = s and the factorization is unique up to associates.

Remark. The converse is not true since Z[x] is a UFD but not a PID.

21 Week 5, Lecture 1 Well-ordering Axiom: Every nonempty subset of the set of nonnegative integers contains a smallest element.

Theorem 5.8. Every Euclidean domain is a PID.

Proof. Let I be an ideal of the Euclidean domain R, and consider the set {δ(a): a ∈ I, a 6= 0}. By The well-ordering principal there is an element b ∈ I such that δ(b) is minimal in the set {δ(a): a ∈ I, a 6= 0}. We claim that hbi = I. So we must show that every a ∈ I is in hbi. Since R is a Euclidean domain, there are elements r and q in R such that b = aq + r and either r = 0 or δ(r) < δ(a). Note that b − aq = r, and also b and a are in I, therefore, b − aq ∈ I. So, r ∈ I. It is not possible to have r 6= 0, since δ(r) < δ(a) and δ(a) is the smallest element in I. So we must have r = 0, and b = aq, i.e., a ∈ hbi.

Deﬁnition. 1.A Dedekind domain is an integral domain in which every nonzero proper ideal factors into a product of prime ideals.

Remark. Let R be the following ring of the complex numbers: √ R = {a + b(1 + 19i)/2 : a, b ∈ Z}. Then R is a principal ideal domain that is not a Euclidean domain.

Deﬁnition. Let A be a nonempty subset of a commutative ring R.An element d is a great common divisor of X provided:

1. d|x for all x ∈ X.

2. if c|x for all x in X, then d|c.

Remark. There are some commutative rings that a set X of its elements does not have a GCD, for example look at the ring 2Z and the set {2, 4}.

Deﬁnition. Let a1, a2, . . . , an be some elements in a ring R with identity. Then if the GCD of a1, a2, . . . , an is 1, then a1, a2, . . . , an are said to be relatively prime.

Theorem 5.9. If R is a UFD, then there is a GCD of a1, a2, . . . , an in R.

mi1 mik Proof. Factor each ai = uip1 . . . pk into irreducible elements, where all pij are distinct k1 kn and mij ≥ 0. Show that d = p1 . . . pn is the greatest common multiple of a1, a2, . . . , an, where kj = min{m1j, . . . , mnj}. √ 6 Z[ d], an integral domain which is not a UFD

An square-free element in Z is an element d 6= 1 such that d = −1 or d = p1p2 . . . pk for distinct prime numbers p1, p2, . . . , pk. For a square-free number d deﬁne √ √ Z[ d] = {a + bZ[ d]: a, b ∈ Z}.

22 √ √ √ Deﬁnition. The function N(s + t d) = (s + t d)(s − t d) = s2 − dt2 is called the norm. √ Theorem 6.1. If d is a square-free integer, then for all a, b ∈ Z[ d] 1. N(a) = 0 if and only if a = 0.

2. N(ab) = N(a)N(b).

Proof. The second√ part is a straight forward computation so we only proof the ﬁrst part. Let a = s + t d. If d = −1, then N(a) = 0 if and only if s2 − dt2 = 0 if and only if s2 = −t2. So we must have s = t = 0. Now assume that d 6= −1. Note that s, t 6= ±1. Then N(a) = 0 if and only if s2 = dt2. Factor s and t into primes. If p be a prime which appears in the factor of d into primes, then in the left hand side p has an even power while in the right hand side it has a odd power, which is impossible. Therefore, the only case we must have is that a = 0 and b = 0.

23 Week 5, Lecture 2 √ Theorem 6.2. Let d be a square-free integer. Then u ∈ Z d is a unit if and only if N(u) = ±1.

Proof. Assume that u is a unit, then there is an element v such that uv = 1, then δ(uv) = 1 and so δ(u)δ(v) = 1 by the previous theorem. Therefore, we must have δ(u) = 1 or −1. √ 2 2 Conversely, assume√ that u = a + b d and δ(u) = 1, thus a − db = 1. Now consider the element v = a − b d, then uv = a2 − db2 = 1. √ √ Example 6.3. Is u = 3 + 2 2 is a unit in Z[ 2]? Yes because δ(u) = 32 − 2.22 = 1. √ Corollary 6.4. Let d be a square free integer. Then if d > 1, then Z[ d] has inﬁnitely many units. If d = −1, then the units are ±1, ±i. If d < −1, then the units are ±1.

Theorem√ 6.5. Let d be a square-free integer. Then every nonzero, nonunit element in Z[ d] is a product of irreducible elements. √ Proof. Let S be the set of all elements in Z[ d] that cannot be written as the product of irreducible elements. We want to show that S = ∅. Assume otherwise, then by well-ordering axiom, the set {|N(t)| : t ∈ S} has an element a such that for any t ∈ S, |N(a)| ≤ N(t). Since a is not irreducible, there are nonunits (because of non-irreduciblity of a) elements b, c such that a = bc and b or c is in S. Without loss of generality assume that b ∈ S. Then N(a) = N(bc) = N(b)N(c) by the previous theorem. Therefore, |N(a)| = |N(b)||N(c)|. If N(b),N(c) = ±1, otherwise they are units which is not possible. Therefore, N(a) > N(b), a contradiction. √ √ Example 6.6. Consider Z[−5]. Then 2 = 2.3 and also 2 = (1 + −5)(1 − −5).

7 The ﬁeld of quotients of an integral domain

Let R be an integral domain. Deﬁne a relation v on the set S = {(a, b): a, b ∈ R, b 6= 0} by (a, b) v (c, d) if and only if ad = bc.

Theorem 7.1. The relation v is an equivalence relation on S.

Proof. Reﬂexive: it is easy to see that (a, b) v (a, b) since ab = ba. Symmetric: if (a, b) v (c, d), then ad = bc, by commutativity cb = da, thus (c, d) v (a, b). Transitivity: if (a, b) v (c, d) v (e, f), then ad = bc and cf = de. Multiplying ad = bc by f we have adf = bcf → a(df) = b(cf) = b(de). Therefore, d(af) = d(be). By cancellation law, we have af = be and so (a, b) = (e, f).

a a c Denote the equivalence class of (a, b), i.e., [(a, b)], by b . Therefore, b = d if and only if ad = bc.

24 Theorem 7.2. Let R be an integral domain. Then the set a F = { , a, b ∈ R, b 6= 0} b is a ﬁeld with the following addition and multiplication,

a c (ad + bc) a c ac + = = . b d (bd) b d bd

Proof. First we showed show that the addition and multiplication are well-defined and then we check that F is a field. 0 0 0 0 a a c c a c a c • Addition is well-defined: Let = 0 and = 0 . We want to show that + = 0 + 0 . b b d d b d b d 0 0 a a c c Since = 0 and = 0 , we have b b d d

ab0 = a0 b cd0 = c0 d.

Therefore, ab0 dd0 = a0 bdd0 cd0 bb0 = c0 dbb0 . So ab0 dd0 + cd0 bb0 = a0 bdd0 + c0 dbb0 Thus we have

0 0 0 0 0 0 0 0 0 0 (ad + bc) (a d + b c ) ⇒ (ad + cd)b d = (a d + c b )bd ⇒ = . (bd) (b0 d0 )

0 0 a a c c • Multiplication is well-deﬁned: Let = 0 and = 0 . We have b b d d

ab0 = a0 b cd0 = c0 d.

0 0 a c a c 0 0 0 0 We want to show that = 0 0 , i.e., (ac)(b d ) = bd(a c ). Note that b d b d

(ac)(b0 d0 ) = (ab0 )(cd0 ) = (a0 b)(c0 d) = (bd)(a0 c0 ).

0 0 a c a c Therefore, = 0 0 . b d b d Moreover, it is straight forward to check that

0 • 0F = b for any b 6= 0. a −a • b + b = 0F . 1 • The identity element is 1 . a b • The inverse of a nonzero element b is a . We have • a c r a c a r ( + ) = + . b d s b d b s

25 Week 5, Lecture 3

8 If R is a UDF, so is R[x]

Deﬁnition. Let R be a UFD. The polynomial f(x) ∈ R[x] is called primitive if the only divisors of f(x) of degree zero are units. Let R be a UFD. 1. The units of R[x] are the units of R. 2. If p is irreducible in R, then it is irreducible in R[x]. 3. An irreducible polynomial is a primitive polynomial. 4. Every polynomial f(x) ∈ R[x] factors as f(x) = cg(x) where c ∈ R and g(x) is a primitive polynomial. 5. Every primitive polynomial of degree 1 is irreducible. Theorem 8.1. Let R be a UFD. Then every nonzero nonunit polynomial f(x) ∈ R[x] is a product of irreducible elements. Proof. We prove this theorem by induction on the degree of f(x). If deg(f(x)) = 0, then f(x) is an irreducible elements of R and since R is a UFD and every irreducible element of R is an irreducible element of R[x], so we have that f(x) is irreducible in R[x]. If deg(f(x)) = 1, then we can write f(x) = cg(x) where c ∈ R and g(x) is a primitive element of degree 1. Since c can be written as product of irreducible elements of R and g(x) is an irreducible element, so f(x) can be written as the product of irreducible elements. Now, assume that deg(f(x)) = r > 1 and for every polynomial of degree less than r the theorem is true. If f(x) is irreducible we are done, otherwise there is a primitive polynomial g(x) such that f(x) = cg(x) where c ∈ R. If g(x) is irreducible then since c can be written as product of primes, f(x) = cg(x) is a product of irreducible elements. If g(x) is not an irreducible element then there are polynomials h(x) and g(x) such that non of them is of degree 0 and g(x) = h(x)k(x). Note that deg(f(x)) = deg(g(x)) > deg(h(x)), deg(k(x)), thus h(x) and k(x) are product of irreducible elements. Therefore, f(x) = ch(x)k(x) is a product of irreducible elements. Lemma 8.2. Let p be an irreducible element in a UFD R. If p|f(x)g(x) where f(x), g(x) ∈ R[x], then p|f(x) or p|g(x). Corollary 8.3. Let R be a UFD. The product of primitive polynomials in R[x] is primitive. Proof. Let f(x), g(x) be primitive polynomials in R[x]. Then if c|f(x)g(x) and c is not a unit and nonzero, then c = p1 . . . pk, where each pi is an irreducible elements. Thus p1|f(x)g(x) and so p|f(x) or p|g(x) which means f(x) or g(x) is not primitive. Theorem 8.4. Let R be a UFD and r, s two nonzero elements of R. Let f(x) and g(x) be primitive elements in R[x] such that rf(x) = sg(x). Then r, s are associates in R and f(x), g(x) are associates in R[x].

26 Proof. If r is a unit then f(x) = r−1sg(x), and since f(x) is primitive, r−1s is a unit and so f(x) and g(x) are associates. If r is not a unit, then it is a product of irreducible elements, so r = p1 . . . pk where pi’s are irreducible. Therefore, p1 . . . pkf(x) = sg(x). We have By lemma 8.2 that p1|sg(x) so p1|s or p1|g(x). Note that g(x) is a primitive element, therefore, p1|s. Let s = p1t. Then p1p2 . . . pkf(x) = p1tg(x). By cancellation law, we have p2 . . . pkf(x) = tg(x). Now p2|tg(x), and similarly we have p2|t. Let t = p2z. Then s = p1p2z. Then p2 . . . pkf(x) = p2zg(x). By cancellation we have p3 . . . pkf(x) = zg(x). Repeat the argument k times, then we have f(x) = wg(x) for some w ∈ R. Since f(x) is primitive we must have w is a unit, and so f(x) and g(x) are associate. Moreover, we have r = p1p2 . . . pk and s = p1p2 . . . pkw for some unit w ∈ R. Therefore, r and s are associate in R. Corollary 8.5. Let R be a UFD, F its ﬁeld of quotients. Let f(x) and g(x) be primitive in R[x]. If f(x) and g(x) are associate in F [x], then they are associate in R[x].

Proof. Since f(x) and g(x) are associate in F [x], there is a unit a/b, a, b ∈ R, b 6= 0 such that f(x) = a/bg(x), then we have bf(x) = ag(x). By the previous theorem we have f(x) and g(x) are associate in R[x].

Corollary 8.6. Let R be a UFD and F be its quotient ﬁeld. If f(x) ∈ R[x] has positive degree and is irreducible in R[x], then f(x) is irreducible in F [x].

Proof. Assume on the contrary that f(x) is not irreducible in F [x]. Then there are 0 0 0 n 0 0 0 n polynomials g(x) = b0/b0 +b1/b1x+...+bn/bnx and h(x) = c0/c0 +c1/c1x+...+cn/cnx in F [x] with positive degree such that f(x) = g(x)h(x). Let b = lcm(b0, . . . , bn). Then bg(x) is in R[x], so there is an element a ∈ R and a primitive polynomial g1(x) ∈ R[x] such that bg(x) = ag1(x). Similarly, we have ch(x) = dh1(x) where c, d ∈ R and h1(x) ∈ R[x] is a primitive polynomial. We have bdf(x) = bdg(x)h(x) = abg1(x)h1(x). By Corollary 8.3, we have g1(x)h1(x) is primitive and also f(x) is primitive too. Therefore, bd and ab are associate in R, and so there is a unit u such that ubd = ab. We have bdf(x) = ubdg(x). By cancellation we have f(x) = ug1(x)h1(x), and so f(x) is not irreducible in R[x]. Theorem 8.7. If R is a UFD, so is R[x].

Proof. We already showed that every polynomial in R[x] is a product of irreducible polynomials. So we must show that this factorization is unique up to reordering and associ- ation. Assume that we factor a nonzero and nonunit polynomial into the following two factors

c1 ··· cmp1(x) . . . pk(x) = d1 ··· dnq1(x) . . . qt(x),

where ci, dj are irreducible in R and pi(x) and qj are irreducible in R[x]. By Theorem 8.4 we have c1 ··· cm and d1 ··· dn are associate in R and also p1(x) . . . pk(x) and q1(x) . . . qt(x) are associates in R[x]. Since R is a UFD and we have c1 ··· cm = (ud1) ··· dn for some unit u ∈ R. We have that m = n and ci = uidi for some units ui and for all i. Let F be the ﬁeld of quotients of R. Then by Corollary 8.6, pi(x) and qj(x) are irreducible in F [x]. Since there is a unit v ∈ R such that p1(x) . . . pk(x) = vq1(x) . . . qt(x) and F [x] is a UFD, we have k = t and after reordering we have that pi(x) and qi(x) are associate in F [x] which by Corollary 8.5 we have thery are associate in R[x].

27 Method of proofs Our goal is to prove a statement. Assume we have a statement P , note that P can be of form A ⇒ B (i.e., an implication),√ for example: if x is odd, then x + 1 is even, or just a proposition A, for example: 2 is irrational. If we want to show that A ⇒ B it is equivalent to show one of the following: Contrapositive: ¬B ⇒ ¬A. Contradiction: (¬B and A) ⇒ C, where C is obviously false. If we want to show that a statement which is just a proposition A, is true, then we can not use contrapositive, and we only can directly prove it or use contradiction: ¬A ⇒ C, where C is obviously false.

9 Vector Spaces

Let F be a ﬁeld. We call (V, +,.) a vector space over F when (V, +) is an abelian group, and . : F × V → V such that for all a, a1, a2 ∈ F and v, v1, v2 ∈ V

1. a(v1 + v2) = av1 + av2;

2.( a1 + a2)v = a1v + a2v;

3. a1(a2v) = (a1a2)v;

4.1 F v = v. Example 9.1. If F and K are fields such that F ⊆ K, we say K is an extension field of F . Any extension field of F is a vector space over F with the same addition as field K, and the scalar multiplication is the multiplication of K.

Let v1, . . . , vn be in the vector space V over F .

• We say vector w is a llinear combination of vectors v1, . . . , vn if there are scalars c1, . . . , cn ∈ F such that w = c1v1 + . . . , cnvn.

• The set of all linear combination of vectors v1, . . . , vn is denoted by Span{v1, . . . , vn}. i.e., Span{v1, . . . , vn} = {c1v1 + ··· + cnvn : c1, . . . , cn ∈ F }.

• We say the set of vectors {v1, . . . , vn} spans V if V = Span{v1, . . . , vn}.

• A subset {v1, . . . , vn} of V is said to be linearly independent provided that whenever c1v1 + . . . , cnvn = 0,

with each ci ∈ F , then c1 = c2 = . . . , ck = 0.

• A subset {v1, . . . , vn} of V is said to be a basis for V if the set is linearly independent and also Span{v1, . . . , vn} = V.

28 Theorem 9.2. Let V be a vector space over a ﬁeld F .

1. The subset {u1, u2, . . . , un} of V is linearly dependent over F if and only if some uk is a linear combination of the preceding ones, u1, u2, . . . , uk−1. 2. All bases for V have the same size (cardinality).

3. If a basis for V has a finite number of vectors, then V is called finite dimensional. The number of elements in any basis of V is called dimension of V , is denoted by [V : F ]. If V does not have a finite basis, then V is said to be infinite dimensional over F .

4. Let {v1, v2, . . . , vk} be a basis for V . If a subset {u1, u2, . . . , un} spans V , then n ≥ k. If a subset {w1, w2, . . . , wm} is linearly independent, then m ≤ k. 5. Let K and F be ﬁelds such that F ⊆ K. Then [K : F ] = 1 if and only if K = F .

Definition. We say that K is a finite-dimensional extension of F if K, considered as a vector space over F , is finite dimensional over F .

Let V and W be vector spaces on a ﬁeld F .A homomorphism f form V to W is a map that for all c ∈ F, v, w ∈ V , f(cv + w) = cf(v) + w. An isomorphism form V to W is a homomorphism that is injective and surjective.

Theorem 9.3. 1. Let F,K and L be fields with F ⊆ K ⊆ L. If [K : F ] and [L : K] are finite, then L is a finite-dimensional extension of F and [L : F ] = [L : K][K : F ].

2. Let K and L be ﬁnite dimensional extension ﬁelds of F and let f : K → L be an isomorphism such that f(c) = c for every c ∈ F . Then [K : F ] = [L : F ].

29 10 Simple Extensions

Let K be an extension of the ﬁeld F and u ∈ K. Let F (u) be the intersection of all subﬁeld of K containing F and u.

• F (u) is a subﬁeld of K. F (u) is the smallest subﬁeld of K containing F and u.

• F (u) is called a simple extension of F . Definition. Let field K be an extension of field F and u ∈ K. Then u is said to algebraic over F if u is the root of some nonzero polynomial in F [x]. When u is not a root of some polynomial we say u is transcendental.

√Example 10.1. i ∈ C is algebraic over R. 2 is algebraic over Q. Theorem 10.2. Let u ∈ K be algebraic over F . Then there exists a unique monic irreducible polynomial p(x) in F [x] that has u as a root. Furthermore, if u is a root of g(x) ∈ F [x], then p(x) divides g(x). Proof. Let S be the set of all nonzero polynomials in F [x] that have u as a root. Since u is algebraic over F , at least there is a polynomial in F [x] that has u as a root, so S 6= ∅. By the Axiom of Choice there is an element p(x) in S with the smallest degree. We now show that p(x) is irreducible. Assume p(x) = f(x)g(x). If both f(x) and g(x) are not constant, then p(u) = f(u)g(u) = 0, and since F is an integral domain, we must have f(u) = 0 or g(u) = 0. Without loss of generality assume that f(u) = 0. However, deg(f(x)) < deg(p(x)), and f(u) = 0, this yields a contradiction since we chose p(x) in a way that it has the smallest degree in S. If a polynomial has u as a root, all constant multiple of that polynomial has u as a root, we may assume that p(x) is monic. Now we show if g(u) = 0 for some polynomial g(x) ∈ F [x], then p(x)|g(x). By division algorithm there are polynomials f(x) and r(x) such that g(x) = p(x)f(x) + r(x) and degree of r(x) is either zero or deg(r(x)) < deg(p(x)). Since g(u) = 0 and f(u) = 0 and we have g(u) = p(u)f(u) + r(u), we have r(u) = 0. Note that r(x) must be zero because we have chosen p(x) in a way that has the smallest degree amongst all polynomials with u as a root. Thereofore, p(x)|g(x). Now we prove that p(x) is unique. Assume p1(x) is also a monic polynomial has u as a root and if u is a root of g(x) ∈ F [x], then p1(x) divides g(x). Therefore, p1(x)|p(x), and since they are irreducible we must have p1(x) = up(x) for some unit u. Since both polynomials are monic we conclude that p(x) = p1(x). Definition. If K is an extension of field of F and u ∈ K is algebraic over F , then the unique monic, irreducible polynomial p(x) in the above thorem is called minimal polynomial of u over F . √ As an example x2 − 3 is the minimal polynomial of 3 over Q. Theorem 10.3. Let K be an extension field of F and u ∈ K be algebraic over F with minimal polynomial p(x) of degree n. Then

1. F (u) ∼= F [x]/hp(x)i.

2 n−1 2. {1F , u, u , . . . , u } is a basis of the vector space F (u) over F . 3. [F (u): F ] = n.

30 Proof. Proof of (1): Define a fucntion ψ : F [x] → F (u) such that ψ(f(x)) = f(u). This function is a ring homomorphism. By the first isomorphism theorem we have Imgψ ∼= F (x)/ker(ψ). So we only need to show that ker(ψ) = hp(x)i and Imgψ = F (u). Note that ker(ψ) = {f(x): f(u) = 0} i.e., the set of all polynomials that have u as a root. By the previous theorem, if a polynomial has u as a root then p(x) divides it, therefore, ker(ψ) ⊆ hp(x)i, and it is clear that hp(x)iker(ψ). Thus hp(x)i = ker(ψ). So F [x]/hp(x)i = Imgψ. Moreover, since image of ψ is a field that contains both F and u it must be F (u). 2 n−1 Proof of (2) and (3): We first show that {1F , u, u , . . . , u } spans F (u). Any element of F (u) is of the form f(u) for some polynomial f(x) ∈ F (x). If deg(p(x)) = n, then by the n−1 devision algorithm we have f(x) = p(x)q(x)+r(x), where r(x) = b0 +b1x+...+bn−1x n−1 for some bi ∈ F . Then f(u) = p(u)q(u) + r(u) = b0 + b1u + ... + bn−1u . 2 n−1 Now we show that {1F , u, u , . . . , u } is linearly independent. Assume on the contrary that the set is not linearly independent, then there are elements a0, a1, . . . , an−1 ∈ F n−1 such that at leat one of them is not zero and a0 + a1u + ... + an−1u = 0. Therefore n−1 the polynomial g(x) = a0 + a1x + ... + an−1x has u as its root, but it is not possible sice p(x) has the smallest degree amongst all polynomials that have u as a root. √ √ √ ∼ 2 Example 10.4. The set {1, 3} is a basis for Q[ 3] and moreover, Q[ 3] = Q[x]/hx − 3i. Corollary 10.5. If u and v have the same minimal polynomial p(x) in F [x], then F (u) ∼= F [v]. Definition. Let R,S be rings and Q and T be subring of R and S respectively. We say the isomorphism σ : Q → T extends to the isomorphism f : R → S if f(r) = σ(r) for every r ∈ Q. Example 10.6. If σ : F → E is an isomorphism of fields, then it extends to (by an abuse of notaiton) σ : F [x] → E[x] n n a0 + a1x + ... + anx 7→ σ(a0) + σ(a1)x + ... + σ(an)x Corollary 10.7. Let σ : F → E be an isomorphism of fields. Let u be an algebraic in some extension of F with minimal polynomial p(x) and v be an algebraic element in some externison of E with minimal polynomial σ(p(x)). Then σ extends to an isomorphism of fields σ : F (u) → F (v) such that σ(u) = v and also for every c ∈ F , we have σ(c) = σ(c). Proof. Note that by the previous theorem τ : E[x]/hσ(p(x))i → E(v) f(x) + hσ(p(x))i 7→ f(v) is an isomorphism. Also define the surjective homomorphism π : E[x] → E[x]/hσ(p(x))i f(x)i 7→ f(x) + hσ(p(x))i Now consider the following map,

ψ : F [x] →σ E[x] →π E[x]/hσ(p(x))i →τ E(v) f(x) 7→ σ(f(x)) 7→ σ(f(x)) + hσ(p(x))i 7→ σ(f(v))

31 We use the ﬁrst isomorphism theorem and we have F [x]/kerψ ∼= E(v) since all maps are surjective. We claim that kerψ = hp(x)i. If f(x) ∈ kerψ, then σ(f(v)) = 0. As τ is an isomorphism, we must have σ(f(x)) ∈ hσ(p(x))i, which is equivalent to say that f(x) ∈ hp(x)i. Moreover, ψ(p(x)) = σ(p(v)) = 0. Therefore, kerψ = hp(x)i. Then ψ : F [x]/hp(x)i → E(v) f(x) + hp(x)i 7→ f(v) is an isomorphism. Also by the previous theorem we have the following isomorphism

−1 φ : F [u] → F [x]/hp(x)i f(u) 7→ f(x) + hp(x)i

Now deﬁne −1 σ = ψ ◦ φ : F [u] → F [v] f(u) 7→ f(v) is an isomorphism such that σ(u) = v and also σ(c) = σ(c) for all c ∈ F .

32 11 Algebraic Extensions

Definition. An extension field K of a field F is said to be an algebraic extension of F if every element of K is algebraic over F .

Example 11.1. The complex number C is an algebraic extension of R. Note that for every element a + ib ∈ C we have (x − (a + bi))(x − ((a − bi))) = x2 − 2ax + (a2 + b2) ∈ mathbbR[x]. Therefore, a + ib is a root of a polynomial in R[x]. Theorem 11.2. If K is a ﬁnite-dimensional extension of F , then K is an algebraic extension of F .

Proof. Let [K : F ] = n and u ∈ K. We have either for some 0 ≤ i < j, ui = uj or all different powers of u are distinct. In the former case, we have u is a root of xi −xj ∈ F [x], 2 n and in the latter case the set {1F , u, u , . . . , u } is liearly dependent since the set contains n + 1 element, therefore, there are scalars c0, . . . , cn ∈ F such that at least one of them n n is nonzero and c0 + c1u + ... + cnu = 0. It follows that c0 + c1x + ... + cnx ∈ F [x] has u as a root. • The inverse of the above theroem is false since there is an algebraic extension over Q with infinite dimestion. See Exercise 16 of Section 11.3. Example 11.3. When u is an algebraic element over F , the field F (u) is algebraic over F since by the previous theorem it is finite dimensional over F .

Deﬁnition. 1. Let F (u1, . . . , ut) be the intersection of all ﬁelds that contains all ui and also F .

2. F (u1, . . . , ut) is said to be a finitely generated extension of F , generated by u1, . . . , ut. √ Example√ 11.4. 1. The field Q( 3, i) is the smallest subfield of C contains Q and both 3 and i.

2. The finitely generated Q(i, −i) is the same as Q(i) and so it is a simple extension. 3. A finite dimensional extension K of a field F is also finitely genetrated since if u1, u2, . . . , uk is a basis for the extension K, then K = (u1, u2, . . . , uk). Remark. Let u, v be two elements in an extension of F , then F (u, v) = F (u)(v).

Proof. F (u, v) contains F (u) since F (u) is a subfield of any field containing both F and u. Moreover, F (u)(v) is the subfiled of any field containing F (u) and v, and so is a subfield of F (u, v). Thus, F (u)(v) ⊆ F (u, v). Also, since F (u, v) is the smallest subfield of containing u, v and F , it is a subfield of F (u)(v). F ⊆ F (u) ⊆ F (u, v) = F (u)(v)

√ Example 11.5. Find the dimestion of Q( 3, i) over Q.

33 √ √ Proof. Note that the monimal polynomial of 3 over Q is x2 − 3, therefore, Q( 3) has 2 dimension√ 2 over Q. Moreover, the minimal polynomial of i over Q is x + 1. Thus, [Q(i): Q( 3)] = 2. Therefore, √ √ [Q( 3, i): Q] = [Q(i): Q( 3)][Q(sqrt3), Q] = 4.

Theorem 11.6. Let K = F (u1, . . . , un) is a ﬁnitely generated extension of F and each ui is algebraic over F , then K is ﬁnite dimensional algebraic extension of F . Proof. Note that we have

F ⊆ F (u1) ⊆ F (u1, u2) ⊆ F (u1, . . . , un)

Thus [F (u1, . . . , un): F ] = [F (u1, . . . , un): F (u1, . . . , un−1)]

[F (u1, . . . , un−1): F (u1, . . . , un−2)] ... [F (u1, u2): F (u1)][F (u1): F ]

For each i,[F (u1, . . . , ui−1): F (u1, . . . , ui)] is the same as [F (u1, . . . , ui−1)(ui): F (u1, . . . , ui−1)], and since ui is algebraic over F , it is also algebraic over F (u1, . . . , ui−1), therefore, by a theorem [F (u1, . . . , ui−1): F (u1, . . . , ui)] is finite and so [F (u1, . . . , un): F ] is finite, and since any finite-dimensional extension is algebraic, we conclude that K = F (u1, . . . , un) is finite dimensional algebraic extension of F . Corollary 11.7. If L is an algebraic extension of K and K is an algebraic extension of F , then L is an algebraic extension of F .

Proof. Let u be an element of L, we should show that there is a polynomial in F [x] that has u as a root. As L is an algebraic extension of K there is a polynomial a0 +a1x+...+ n n anx such that a0 + a1u + ... + anu = 0. Note that u is algebraic over F (a0, . . . , an). Also, note that F (a0, . . . , an, u) is ﬁnite dimensional over F . Therefore, F (a0, . . . , an, u) is an algebraic extension of F and so there is a polynomial in F [x] that has u as a root.

Corollary 11.8. Let K be an extension field of F and let E be the set of all elements of K that are algebraic over F . Then E is a subfield of K and an algebraic extension field of F .

Proof. It is clear that if E is a field, it is an algebraic extension of F , so we only need to show that it is a subfield of K. If u, v ∈ E, then F (u, v) is a subset of E since all elements of F (u, v) are algebraic over F . Therefore, we have u+v, −v, uv ∈ F (u, v) ⊆ E, and moreover, u−1 ∈ F (u, v) if u is not zero. Therefore, E is a field.

Deﬁnition. The set of all elements of C that are algebraic over Q is called the ﬁeld of algebraic numbers.

34 12 Splitting Fields

Any polynomial of degree n over a ﬁeld has at most n roots.

Deﬁnition. We say that f(x) split over the ﬁeld K if f(x) factors in K[x] as

f(x) = c(x − u1) ... (x − un).

Definition. If F is a field and f(x) ∈ F [x], then an extension field K of F is said to be splitting field (or root field) of f(x) over F provided that

1. f(x) splits over K, say f(x) = c(x − u1) ... (x − un).

2. K = F (u1, . . . , un).

Example 12.1. The splitting field of x2 + 1 over R is R(i, −i) = R(i) = C. Lemma 12.2. Let f(x) ∈ F [x] where F is a field. Then there is an extension field of F that contains a root of f(x).

Proof. As F [x] is a UFD, then f(x) factors into irreducible polynomials. So assume that n f(x) = cp(x)p1(x) . . . pk(x) where all pi(x) are irreducible and p(x) = a0 + a1x + ... + x is a monic irreducible polynomial. So if we show that there is an extension of F that contains a root of p(x), then it is also contains a root of f(x). Note that K = F [x]/hp(x)i is an extension of F . Moreover, consider x + hp(x)i. Then

n n p(x + hp(x)i) = a0 + a1(x + hp(x)i) + ... + (x + hp(x)i) = a0 + a1x + ... + anx =

p(x) + hp(x)i) = 0K .

Theorem 12.3. Let F be a ﬁeld and f(x) a nonconstant polynomial of degree n in F [x]. Then there exists a splitting ﬁeld K of f(x) over F such that [K : F ] ≤ n!.

Proof. We proceed the proof by induction on degree of f(x). If degree of f(x) is one then F is a splitting field for f(x) and so [F,F ] = 1 ≤ 1!. Now assume that for every polynomial of degree n − 1 the theorem is true, i.e., any nonconctant polynomial over any field of degree n − 1 has a splitting field. By previous lemma there is an extension field K that contains a root, say u, of f(x). Therefore, f(x) = c(x − u)g(x) where g(x) ∈ F (u)[x]. By induction hypothesis, there is a splitting field K of g(x) over F (u) such that [K : F (u)] ≤ (n − 1)!. We have also that K is splitting field of f(x) over F , and [K : F ] = [K : F (u)][F (u): F ] ≤ (n − 1)!deg(f(x)) ≤ n!.

Any two splitting ﬁeld of a polynomial in F [x] are isomorphic.

Theorem 12.4. Let σ : F → E be an isomorphism of ﬁelds. Assume that f(x) ∈ F [x] is nonconstant. Let σ : F [x] → E[x] f(x) 7→ σ(f(x))

• K a splitting ﬁeld of f(x) over F

• L a splitting ﬁeld of σ(f(x))

35 Then σ extends to an isomorphism between K and L.

Proof. We proceed the proof by induction on deg(f(x)). If deg(f(x)) = 1, then f(x) = c(x − b) where c and b are elements of F . Therefore, the splitting ﬁeld of f(x) is itself. Also σ(c(x − b) = σ(c)x + σ(cb) = σ(c)(x − σ(b)), since σ(c), σ(b) ∈ E, we have that splitting ﬁeld of σ(f(x)) is E, and we already have that F ∼= E. Assume that the theorem is true for any polynomial of degree n−1 and deg(f(x)) = n. Assume that u ∈ K is a root of f(x) and p(x) is the minimal polynomial of u. Consider that f(x) ∈ F (u)[x]. Use division algorithm to divide f(x) by x − u in F (u)[x]. Then we have f(x) = (x − u)g(x) for some g(x) ∈ F (u)[x]. Consider that σ(p(x)) is a monic irreducible polynomial. Let v ve a root of σ(P (x)). So σ(P (x)) is the minimal polynomial of v. By a theorem we have the isomorphism σ : F → E extends to an isomorphism from F (u) to E(v). Now we have

K L

⊆ ∼ ⊆ F (u) →= E(v) ⊆ ⊆ F →σ E

If f(x) = (x − u)(x − u1) ... (x − uk) then g(x) = (x − u1) ... (x − uk). Note that g(x) ∈ F (u)[x] has splitting ﬁeld K, moreover,

σ(f(x)) = σ((x − u)g(x)) = (x − σ(u))σ(g(x)) = (x − v)σ(g(x)).

So the splitting ﬁeld of σ(g(x)) ∈ E(v) is also L. Therefore, by induction hypothesis the isomorphism between F (u) and E(v) extends to an isomorphism between K and L.

Deﬁnition. An algebraic extension ﬁeld K of F is normal provided that whenever an irreducible polynomial in F [x] has one root in K, then it splits over K.

Theorem 12.5. The field K is a splitting field over the field F of some polynomial in F [x] if and only if K is finite dimensional, normal extension of F .

Proof. As K is a splitting ﬁled of some polynomial f(x) ∈ F [x], we have K = F (u1, . . . , un) where ui are roots of f(x). Since each ui is algebraic over F , we have from a theorem that K is ﬁnite dimensional extension of F .

Now we show that K is normal extension of F . Let p(x) be an irreducible polynomial in F [x] with a root u in K. We want to show that if w is a root of p(x) other than u, then w ∈ K. Consider the ﬁeld extension F (w). Since w and u has the same minimal polynomial we have that F (u) ∼= F (w). So we have K K(w) ⊆ ⊆ F (u) ∼= F (w) ⊆ ⊆ F = F

36 Since K is splitting field of f(x) ∈ F (u)[x] and also K(w) is also a splitting field of f(x) ∈ F (w)[x] and moreover F (u) ∼= F (w), by the last theorem we have K ∼= K(w), in such a way that this isomorphism takes u to w and any element of F maps to itself. By a theorem in linear algebra ( Let K and L be finite dimensional extension fields of F and let f : K → L be an isomorphism such that f(c) = c for every c ∈ F . Then [K : F ] = [L : F ].) we have [K : F ] = [K(w): F ], so K is a subspace of K(w) with the same dimension, and so K(w) = K. Therefore, w ∈ K. Conversely, assume that K is a finite-dimensional normal extension of F , we want to show that there is a polynomial f(x) such that K is its splitting field. Since K is finite dimensional, then K has a basis {u1, . . . , uk} over F , so we can write K = F (u1, . . . , uk). Note that by the theorem [If K is a finite-dimensional extension field of F , then K is an algebraic extension of F ], each ui has a minimal polynomial over F , say pi(x). Since pi(x) has one root, ui, in K and K is normal we conclude that all roots of pi(x) are in K. Therefore, K is the splitting field of f(x), a polynomial over F .

√ Example 12.6. Fact: z = 22/3( −1+ 3i ) is a root of x3 − 2 in . √2 C 3 Use the above fact to show√ that Q( 2) is not a normal extension of Q. 3 Answer: Note that if Q( 2) is a normal extension of Q, then if it has one root of some 3 polynomial of Q[x], it must contains all√ of other roots of the polynomial. But x − 2 is in 3 Q[x] with a root z which is not in Q( 2) Deﬁnition. 1. A ﬁeld over which every non-constant polynomial splits is said to be algebraically closed. For example, C is algebraically closed. 2. If K is an algebraic extension of F and K is algebraically closed, then K is said to be algebraic closure of F .

37 13 Separability

Definition. 1. Let F be a field. A polynomial f(x) ∈ F [x] of degree n is said to be separable if it has n distinct roots in some splitting field. Equivalently, f(x) is separable if it has no repeated roots in any splitting field.

2. If K is an extension ﬁeld of F , then an element u ∈ K is said to be separable over F if u is algebraic over F and its minimal polynomial p(x) ∈ F [x] is separable.

3. The extension ﬁeld K is said to be a separable extension if every element of K is separable over F .

2 n Derivative of polynomial: The derivative of f(x) = c0 +c1x+c2x +...+cnx ∈ F [x] is deﬁned to be the polynomial

0 n−1 f (x) = c1 + 2c2x + ... + ncnx . ∈ F [x].

If you check we have

(f + g)0 (x) = f 0 (x) + g0 (x)(fg)0 (x) = f(x)g0 (x) + f 0 (x)g(x).

Lemma 13.1. Let F be a ﬁeld and f(x) ∈ F [x]. If f(x) and f 0 (x) are relatively prime in F [x], then f(x) is separable.

Proof. Two polynomials are relatively prime if the only factor of both of them is 1. Assume on the contrary that f(x) is not separable, so in the splitting ﬁeld of f(x) we must have f(x) = (x − u)2g(x). So, f 0 (x) = 2(x − u)g(x) + (x − u)2g0 (x). Therefore, we can see that (x−u) divides both f(x) and f 0 (x), and so they are not relatively prime.

Deﬁnition. Let F be a ﬁeld. Then F has characteristic 0 if n1F 6= 0F for every positive integer n, and also F has characteristic k if k is the smallest integer such that k.1 = 0.

Proposition 13.2. If F is a ﬁled then either it has characteristic 0 or p where p is a prime number.

Proof. Let F be a ﬁled with nonzero characteristic. Let n.1 = 0 where n is the smallest positive integer with this property. If n is not a prime then we can write n = mk, so mk1 = 0, which means (m.1)(k.1) = 0, and so m.1 = 0 or k.1 = 0 which is a contradiction since m and k are smaller than m.

Theorem 13.3. Let F be a ﬁeld of characteristic 0. Then every irreducible polynomial in F [x] is separable, and every algebraic extension of F is a separable extension.

Proof. Let p(x) be an irreducible polynomial in F [x]. So p(x) is nonconstant and it is of the form p(x) = cxn + lower terms and also p0 (x) = (nc)xn−1 + lower terms. Therefore, p0 (x) has a smaller degree than p(x) so they are relatively prime and so p(x) is separable. In particular, the minimal polynomial of each u ∈ K is separable and so K is a separable extension.

Lemma 13.4. Let K = F (v, w) be a separable extension of F and F is an inﬁnite ﬁeld, then there is u ∈ K such that F (u) = K.

38 Proof. Let p(x) be the minimal polynomial of v over F and q(x) be the minimal polynomial of w over F . Let L be the splitting ﬁeld of the polynomial p(x)q(x). Let v, v1, . . . , vn be the roots of p(x) and w, w1, . . . , wm be the roots of q(x). Since F (u, v) is a separable extension of F and w ∈ F (u, v), so the minimal polynomial of q(x) is separable and so all w, w1, . . . , wm are distinct.. As F is inﬁnite, there is an element c ∈ F such that v − v c 6= 0 and c 6= i 1 ≤ i ≤ n, 1 ≤ j ≤ m. w − wj

Let u = v + cw (v = u − cw). We claim that F (u) = K. Deﬁne h(x) = p(u − cx) ∈ F (u)[x]. Note that h(w) = p(u − cw) = p(v) = 0. So w is a root of h(x). We show that the only common root of h(x) and q(x) is w. Assume otherwise, then for some wj we have p(u − cwj) = 0, and so u − cwj = vi. Therefore, v + cw = u = vi + cwj which means v − v c = i . w − wj

A contradiction, thus we must have h(x) ∈ F (u)[x] and q(x) ∈ F (u)[x] has one root w in common. Let r(x) be the minimal polynomial of w over F (u). Then r(x)|h(x) and r(x)|q(x), and so it must be of degree 1 because otherwise, h(x) and q(x) have more than one common roots. So r(x) = a(x − w) such that a ∈ F (u) and w ∈ F (u). Since w ∈ F (u) and u = v − cw we have that v ∈ F (u). And so F (v, w) = F (u).

Theorem 13.5. If K is a ﬁnitely generated separable extension ﬁeld of F , then K = F (u) for some u ∈ K.

Proof. As K is ﬁnitely generated we have K = F (u1, . . . , un) for some ui ∈ K. We proceed by induction on n. If n = 1, so nothing to prove. Assume n ≥ 2 and the theorem is true for n − 1. So F (u1, . . . , un) = F (u1, . . . , un−1)un). By induction hypothesis we have there is some v ∈ F (u1, . . . , un−1) such that F (u1, . . . , un−1) = F (v). Therefore, F (u1, . . . , un) = F (u1, . . . , un−1)un) = F (v, un), and so by previous lemma, there is an element u such that F (v, un) = F (u).

39 14 Finite Fields

Let R be a ring with identity. We say R has characteristic 0 if there is not a positive integer m such that m.1 = 0 and we say it has characteristic n if n is the smallest positive integer such that n.1 = 0. Theorem 14.1. If R is an integral domain the characteristic of R is either inﬁnity or a prime number. Lemma 14.2. Let R be a ring with identity of characteristic n > 0. Then k.1 = 0 if and only if n|k. Proof. We can write k = mn + r, we have 0 = k.1 = (mn + r)1 = mn1 + r1 = 0 + r1, so if r is not zero, then r1 = 0 and r < n a contradiction. Therefore, we must have r = 0 and n|k. Theorem 14.3. Let R be a ring with identity. Then

1. The set P = {k1R|k ∈ Z} is a subring of R. ∼ 2. If R has characteristic 0, then P = Z. ∼ 3. If R has characteristic n > 0, P = Zn. Proof. It is easy to check that P is closed under subtraction and also multiplication. Therefore, it is a subring. To prove (2) and (3), deﬁne f : Z → P given by f(k) = k.1. It is clear that f is a surjective homomorphism. If char(R) = 0, then the kernel of f is ∼ trivial and so Z = P , if char(R) = n, then kernel of f is equal to {k.1 = 0 : k ∈ Z}. Since if any element k with k.1 = 0 divides n, we have kerf = nZ and so be ﬁrst isomorphism ∼ theorem Zn = Z/nZ = P .

Corollary 14.4. Every field of characteristic p has Zp as a subfield. Theorem 14.5. Every finite field K has order pn, where p is the characteristic of K and [K : Zp] = n.

Proof. Any finite field has characteristic > 0, therefore, it contains Zp by previous theorem. So it is a vector space over Zp. By linear algebra, when K is finite, we have the dimension of a vector space K over Zp is |K|/|Zp| = n. Theorem 14.6. (Freshman’s Dream) Let p be a prime and R be a commutative ring with identity of characteristic p. Then for every a, b ∈ R and every positive integer n, (a + b)pn = apn + bpn . Proof. We proceed the theorem by induction on n. If n = 1, then by the Binomial Theorem, p p p (a + b)p = ap + ap−1b + ··· + ap−rbr + ··· + abp−1 + bp. 1 r p − 1

p p p−r r The prime p divides each of the coeﬃcients r , so the term r a b = 0. Therefore, (a + b)p = ap + bp. Assume that the theorem is true for n = k. Now by induction hypothesis and ﬁrst step we have (a + b)pn+1 = (a + b)pn p = (apn + bpn )p = apn+1 + bpn+1 .

40 Theorem 14.7. Let K be an extension ﬁeld of Zp and n a positive integer. Then K has n pn order p if and only if K is a splitting ﬁeld of x − x over Zp.

pn Proof. ⇐) As K is a splitting field of x − x over Zp, it contains all of its roots. We show that xpn − x has pn distinct roots, and the set all of these distinct roots is precisely K. Let E be the subset of K containing all of the roots of xpn − x. Note that if f(x) = xpn − x, then f 0 (x) = pnxpn−1 − 1 = −1. Therefore, f(x) and f 0 (x) are relatively prime. And so xpn − x is separable. So E has pn distinct elements. If we show that E is a field, then since E also spitting field of xpn − x we conclude that K = E, and so it has pn elements. Let a, b ∈ E and a 6= 0. Then

(a − b)pn − (a − b) = (a + (−b))pn − (a + (−b)).

By Freshman’s dream

(a + (−b))pn − (a + (−b)) = apn + (−b)pn − (a + (−b)) = (a + (−b)) − (a + (−b)) = 0.

So a − b ∈ E. Moreover,

(ba−1)pn − ba−1 = bpn (a−1)pn ) − ba−1.

Since a, b ∈ E, we have apn − a = 0 and so apn = a and similarly, bpn = b. Therefore,

bpn (a−1)pn ) − ba−1 = ba−1 − ba−1 = 0, so ba−1 ∈ E. We can now say that E is a ﬁeld.

Assume that K is a ﬁeld of order pn. It is enough to show that every element c ∈ pn pn K = {0, c1, . . . , cpn−1} is a root of x − x. If c is zero, then it is a root of x − x. If c 6= 0, then cc1, . . . , ccpn−1 are also the list of all nonzero elements of K. Therefore, u = cc1, ··· , ccpn−1 = c1 . . . cpn−1. Also

pn−1 pn−1 u = cc1, ··· , ccpn−1 = c c1 . . . cn = c u.

Therefore, cpn−1 = 1 which means that cpn = c ⇒ cpn−1 − c = 0, i.e., c is a root of xpn − x.

Corollary 14.8. For each positive prime p and positive integer n, there exists a ﬁeld of order p.

Proof. We previously showed that the splitting ﬁeld of any polynomial exists, and so the splitting ﬁeld of xpn − x exists and by previous theorem has order pn.

Corollary 14.9. Two ﬁnite ﬁelds of the same order are isomorphic.

Proof. Let K and L be two field of order pn. Then they are splitting field of xpn − x, and by Theorem 12.4, they are isomorphic. So there is a unique field, up to isomorphism, of order pn, and we call it Galois field of order pn.

41 Theorem 14.10. Let K be a finite field and F a subfield. Then K is a simple extension of F .

Proof. Note that K \{0} is a multiplicative group, and by a theorem the multiplicative group of any ﬁeld is cyclic. So, there is an element u ∈ K such that K = {1, u, . . . , upn−1}. Therefore, K = F (u).

Corollary 14.11. Let p be a positive prime. For each positive integer n, there exists an irreducible polynomial of degree n in Zp[x].

n Proof. There is a ﬁeld K of order p , and also K has a copy of Zp. By the previous theorem there is an element u ∈ K such that K = Zp(u) so the minimal polynomial of u has degree n.

42 15 The Galois Group

Deﬁnition. Let K be an extension of F . An F -automorphism σ of K is an isomorphism σ : K → K such that for every c ∈ F , σ(c) = c. The set of all F -automorphism is denoted by GalF K and is called the Galois group of K over F .

Theorem 15.1. If K is an extension ﬁeld of F , then GalF K is a group under the operation of composition of functions.

Proof. If σ, τ ∈ GalF K, then σ ◦τ is an isomorphism and also σ ◦τ(c) = σ(τ(c)) = σ(c) = c. Therefore, σ ◦ τ ∈ GalF K. Moreover,

1. Identity map is in GalF K.

−1 −1 2. If σ ∈ GalF K, then σ is an isomorphism such that σ (c) = c because σ is one-to-one and σ(c) = c.

3. Compositions of functions is associative.

Therefore, GalF K is a group. Theorem 15.2. Let K be an extension ﬁeld of F and f(x) ∈ F [x]. If u ∈ K is a root of f(x) and σ ∈ GalF K, then σ(u) is also a root of f(x).

2 n Proof. Let f(x) = a0 + a1x + a2x + ... + anx . Assume that u ∈ K is a root of f. Then

2 n f(σ(u)) = a0 + a1σ(u) + a2σ(u) + ... + anσ(u) .

Note that for every i, σ(u)i = σ(u) . . . σ(u) = σ(un). Therefore,

2 n f(σ(u)) = a0 + a1σ(u) + a2σ(u ) + ... + anσ(u ) =

2 n 2 n σ(a0)+σ(a1)σ(u)+σ(a2)σ(u )+...+σ(an)σ(u ) = σ(a0+a1u+a2u +...+anu ) = σ(0) = 0.

Remark. Let p(x) be an irreducible polynomial in F [x] and K be the splitting ﬁeld of p(x). The above theorem is stating that if u is a root the image of u by any element of GalF K also is a root of f. The converse is true by the following theorem, i.e., the set of all roots of p(x) is {σ(u): σ ∈ GalF K} for any root u of p(x). Theorem 15.3. Let K be the splitting ﬁeld of some polynomial over F and u, v ∈ K. Then there exists σ ∈ GalF K such that σ(u) = v if and only if u and v have the same minimal polynomial in F [x].

Proof. ⇐) Let u and v have the same minimal polynomial p(x), we previously had there is an isomorphism σ : F (u) → F (v) such that σ1(u) = v and σ1 is ﬁxed over F . Consider that K is splitting ﬁeld of some polynomial in F (u)[x] and F (v)[x]. Now, by Theorem 12.4, σ1 extends to an isomorphism σ of K which is the same as σ1 on F (u), i.e., σ(u) = σ1(u) = v. Therefore, σ ∈ GalF K and σ(u) = v. Converse is merely a result of previous theorem.

Example 15.4. Show that GalRC is isomorphic to Z2.

43 Proof. Note that x2 + 1 is an irreducible polynomial in R[x] with roots i and −i. For every τ ∈ GalRC, therefore, we have either τ(i) = i or τ(i) = −i. Therefore, τ(a + ib) = τ(a) + τ(i)τ(b) = a + τ(i)b. Which means τ only can be one of the following automorphisms τ(a + ib) = a + ib τ(a + ib) = a − ib.

Remark. The example above shows that any R-automorphism of C = R(i) is determined by its action on i. This argument is true in general by the following theorem.

Theorem 15.5. Let K = F (u1, . . . , un) be an algebraic extension ﬁeld of F . If σ, τ ∈ GalF K and σ(ui) = τ(ui) for each i = 1, 2, . . . , n, then σ = τ. Proof. To show that σ = τ, it is enough to show that τ −1 ◦ σ = id. We show this by induction on n. Let n = 1. Then K = F (u1), and any element w in K is of the form k −1 −1 w = c0 + c1u1 + ... + cku1, where each ci is in F . Now τ ◦ σ(w) = τ ◦ σ(c0 + c1u1 + k −1 −1 k k ... + cku1) = c0 + c1τ ◦ σ(u1) + ... + ck(τ ◦ σ(u1)) = c0 + c1u1 + ... + cku1 = w. −1 Now assume that for any element in F (u1, . . . , un−1), we have τ ◦ σ = id. Now consider that K = F (u1, . . . , un) = F (u1, . . . , un−1)(un) so any element of K is −1 −1 k of the form w = a0 +a1un +...+akun, and so τ ◦σ(w) = τ ◦σ(a0 +a1un +...+akun). By induction hypothesis we have −1 k −1 −1 k τ ◦ σ(a0 + a1un + ... + akun) = a0 + a1τ ◦ σ(un) + ... + (τ ◦ σ(un)) k = a0 + a1un + ... + akun = w.

√ √ Example 15.6. By Using the above Theorem we want to find√GalQQ√( 3, 5). 2 2 Note that x − 3 and x − 5 are the minimal√ polynomial√ √ of 3 and√ 5 over Q, respec- 2 tively. By Theorem 15.3, for any σ ∈ GalQQ( 3, 5), σ( 3) and σ 5 are roots√ of x√−3 2 and x − 5. Therefore, we have the following possibles situations if σ ∈ GalQQ( 3, 5). √ √ √ √ √ √ √ √ σ( 3) = 3 σ( 3) = − 3 σ( 3) = 3 σ( 3) = − 3  √ √  √ √  √ √  √ √ σ(− 3) = − 3 σ(− 3) = 3 σ(− 3) = − 3 σ(− 3) = 3 √ √ √ √ √ √ √ √ σ( 5) = 5 σ( 5) = 5 σ( 5) = − 5 σ( 5) = − 5  √ √  √ √  √ √  √ √ σ(− 5) = − 5 σ(− 5) = − 5 σ(− 5) = 5 σ(− 5) = 5 √ √ Now√ if√ we show that each of this situations yields an isomorphism√ from√Q( 3, 5) to ∼ Q( 3, 5), and also the order of each one is at most 2, then√Gal√QQ( 3, 5) = Z2 × Z2. For instance we only show that√ the last one√ is in GalQQ( 3, 5). Let ι : Q → Q be 2 identity map. Then since Q( 3) and Q(− 3)√are the splitting√ fields of x −√3 over Q√, so ι can be extend√ to an√ isomorphism σ1 form Q(√ 3)√to Q(− 3) √such√ that σ1( 3) = − 3, and so σ1(− 3) = 3.√ Similarly, since Q( 3)( 5) and Q( 3)( 5) are the√ splitting√ 2 fields√ of√x − 5 over Q( 3), so σ1 extends to an isomorphism σ from Q( 3)( 5) to ( 3)( 5) such that Q √ √ σ( 3) = − 3  √ √ σ(− 3) = 3 √ √ σ( 5) = − 5  √ √ σ(− 5) = 5

44 Also it is easy to check that σ2 = id. Similarly we can show others also are isomorphisms, and it is easy to check the order of each one is at most 2.

Definition. A field E such that F ⊆ E ⊆ K is called an intermediate field of the extension. Note that GalEK ⊆ GalF K.

Theorem 15.7. Let K be an extension ﬁeld of F . Let H be a subgroup of GalF K, and let EH = {k ∈ K : σ(k) = k for every σ ∈ H}.

Then EH is an intermediate ﬁeld of the extension.

−1 Proof. It is enough to show that for every elements a, b ∈ EH , b 6= 0, ab ∈ EH and also −1 −1 −1 a − b ∈ EH . If we want to show that ab ∈ EH we must show that σ(ab ) = ab . Note that for every σ ∈ H,

σ(ab−1) = σ(a)σ(b−1) = σ(a)σ(b)−1 = ab−1

and σ(a − b) = σ(a) − σ(b) = a − b.

Definition. The field EH is called the fixed field of the subgroup H.

Example 15.8. Consider GalRC and let H = GalRC, ﬁnd EH .

Proof. Remember, EH = {k ∈ C : σ(k) = k for every σ ∈ H}. We previously showed that GalRC = {id, τ}. Note that id fixes all elements, and τ only fixes the real part of each complex number, therefore, EH = R. √ √ 3 3 Example 15.9. Consider GalQQ( 2) and let H = GalQQ( 2), find EH . √ √ 3 3 3 Proof. Let σ ∈√ H, then note that σ( 2) = 2. Note√ √ that √x − 2 is the minimal 3 3 3 3 2 √polynomial of 2 and the roots of this polynomial√ are 2, 2w,√ 2w where w = (−1 + 3 3 3 3i)/2. Note that the only real root of x − 2 is 2. Also, Q( 2)√ is only contains real 3 numbers, so√ since√ any σ ∈ H has an image√ in real numbers, and σ( 2) is also a root of 3 3 3 3 x − 2, σ( 2) = 2. Therefore, EH = Q( 2).

16 The Fundamental Theorem of Galois Theory

Throughout this section let K be a ﬁnite dimensional extension ﬁeld of F . Let

S = {E : F ⊆ E ⊆ K} T = {H : H ⊆ GalF K}.

The main goal of this section is to show that for Galois extension K of F ,

ϕ : S → T E 7→ GalEK

is a bijection. Note that ϕ(F ) = GalF K and ϕ(K) = GalK K = id.

45 Lemma 16.1. Let K be a finite dimensional extension field of F . If H ⊆ GalF K, then K is a simple, normal, separable extension of EH (fix field of H), for simplicity we denote EH by E. Proof. We had a theorem that ant finite dimensional extension field is algebraic so K is algebraic extension of F , and since F ⊆ E, K is algebraic extension of E too. We now want to show that K is a separable extension of E. Let u ∈ K and p(x) ∈ E[x] be the minimal polynomial of u. Then consider that {σ(u): σ ∈ H} is a root of p(x), and therefore, this set if a finite set, so let

{σ(u): σ ∈ H} = {u1, u2, . . . , ut}.

Also for every σ ∈ H, σ(u1), σ(u2), . . . , σ(ut) are distinct roots of p(x) because σ is one-to-one. Therefore,

{u1, u2, . . . , ut} = {σ(u1), σ(u2), . . . , σ(ut)}.

Every σ ∈ GalF K extends to an isomorphism from K[x] to K[x] which by abuse of notation was dented by σ. Now, let f(x) = (x − u1)(x − u2) ... (x − ut). We show that f(x) ∈ E[x]. Note that for every σ ∈ H, we have

f(x) = (x − u1)(x − u2) ... (x − ut) = (x − σ(u1))(x − σ(u2)) ... (x − σ(ut)) = σ(f(x)).

t Consider that we can write f(x) = a0 + a1x + ... + atx ∈ K[x]. Since for every σ ∈ H, σ(f(x)) = f(x), we have that for each σ ∈ H and ai, σ(ai) = ai. Therefore, all ai ∈ E, and so f(x) ∈ E. So we showed that for any arbitrary element u ∈ K, it is a root of a separable polynomial over E, so K is a separable extension of E. Moreover, any ﬁnitely generated separable extension is simple so K = E(u) for some u ∈ K (see theorem 13.5). Also since K is splitting ﬁeld of f(x) ∈ E[x], then as a result of Theorem 12.5, K is a normal extension of E.

Theorem 16.2. Let K be a ﬁnite dimensional extension ﬁeld of F . If H ⊆ GalF K, then

H = GalEH K and [K : E] = |H|.

Proof. We ﬁrst show that H ⊆ GalEH K. Let σ ∈ H, then by the deﬁnition of the

ﬁxed ﬁeld EH , for every k ∈ EH , σ(k) = k, therefore, σ ∈ GalEH K. We can say that

H ⊆ GalEH K.

Now, since GalEH K is ﬁnite, if we show that |H| ≥ |GalEH K|, then H = GalEH K. By the previous theorem we have that K = EH (u) for some u ∈ K. Let p(x) be the minimal polynomial of u over EH (x). Let

f(x) = (x − u)(x − u1) ... (x − ut),

where {σ(u): σ ∈ H} = {u, u1, . . . , ut}. If we show that

[K:EH ]=

|H| ≥ deg(f(x)) ≥ deg(p(x)) ≥ |GalEH K| ≥ |H|

then H = GalEH K and [K : EH ] = |H|.

46 For the ﬁrst inequality, it is clear that |H| ≥ |{σ(u): σ ∈ H}| = |{u, u1, . . . , ut}| = deg(f(x)). For the second inequality, same as the previous theorem f(x) ∈ EH [x], and moreover, it has u as a root, so p(x)|f(x), and it follows that deg(f(x)) ≥ deg(p(x)) = [K : EH ].

For the third inequality, note that p(x) is separable, and also for every σ ∈ GalEH K,

σ(u) is a root of p(x). Note that {σ(u): σ ∈ GalEH K} contains exactly |GalEH K| elements, because if σ1(u) = σ2(u), then for every element in K = EH (u), σ1 = σ2. Therefore,

deg(p(x)) ≥ |GalEH K|.

√ 3 Example 16.3. Previously√ we showed that GalQQ(√ 2) = hιi. Here we have two inter- 3 3 mediate ﬁelds Q and Q( 2), but ϕ(Q) = hιi = ϕ(Q( 2)). In this case, ϕ is not injective, so we need more condition on K as an extension of F .

16.1 Galois Extensions Definition. If K be a finite-dimensional, normal, separable (FDNS) extension field of the field F , we say that K is Galois extension of F , or that K is Galois over F

Theorem 16.4. Let K be a Galois extension of F and E an intermediate field. Then E is the fixed field of the subgroup GalEK, i.e., EGalE K = E.

Proof. Note that EGalE K = {k ∈ K : σ(k) = k, ∀σ ∈ GalEK}, therefore, since for every k ∈ E and σ ∈ GalEK, σ(k) = k, we must have E ⊆ EGalE K . So we only need to show that EGalE K ⊆ E. We proceed the proof by contradiction. Assume that there is u ∈ EGalE K \ E. Since K is a separable extension of F , there is an irreducible separable polynomial p(x) ∈ F (x) which has u as a root. Moreover, let q(x) be the minimal polynomial of u over E[x]. Note that q(x)|p(x) and since p(x) ∈ F [x] is separable, we have that q(x) is separable. Moreover, K is normal extension of F and so all of the roots of p(x) and so q(x) are in K. Since K is ﬁnite dimensional over E, it is splitting ﬁeld of some polynomial over E. So by Theorem 15.3, {σ(u): σ ∈ GalEK} is the set of all roots of q(x). Note that q(x) has degree more than one since a root of it, i.e., u is not in E. Assume that v is another root of p(x), then there is a σ ∈ GalEK such that σ(u) = v, which means that u 6∈ EGalE K , a contradiction.

By the previous two theorems for a Galois extension K of F , F ⊆ E ⊆ K, and H ⊆ GalF K, we have

EGalE K = EH = GalEH K.

Corollary 16.5. Let K be a Galois extension of F . Let

S = {E : F ⊆ E ⊆ K} T = {H : H ⊆ GalF K}.

Then ϕ : S → T E 7→ GalEK is a bijection.

47 Proof. Since a Galois extension is ﬁnite dimensional, by Theorem 16.2, for any H ⊆

GalF K, we have H = GalEH K, and so ψ is surjective. Moreover, by the previous theorem if GalE1K = ϕ(E1) = ϕ(E2) = GalE2K, then E = E = E = E . 1 GalE1 K GalE2 K 2 Therefore, ϕ is injective. Corollary 16.6. Let K be a ﬁnite-dimensional extension of F . Then K is Galois over

F if and only if F = EGalF K , i.e., F is the ﬁxed ﬁeld of the Galois group GalF K.

Proof. If K is Galois over F again by the previous theorem we have F = EGalF K . Con- versely, if F = EGalF K , then by Theorem 16.1, K is a normal, simple, separable extension of F and so it is a Galois extension.

√ √ Q( 3, 5) hιi

√ √ √ Q( 3) Q( 5) Q( 15) {ι, α} {ι, τ} {ι, β}

Q {ι, τ, α, β}

Theorem 16.7. (The Fundamental Theorem of Galois Theory) If K is a Galois extension of F , then 1. ϕ is a bijection. Furthermore,

[K : E] = |GalEK| and [E : F ] = [GalF K : GalEK].

2. An intermediate ﬁeld E is a normal extension of F if and only if GalEK C GalF K, and in this case ∼ GalF E = GalF K/GalEK. Proof. (1) We have already showed in Corollary 16.5 that ϕ is a bijection. Note that by

Theorem 16.4 EGalE K = E, therefore, by Theorem 16.2,

[K : E] = [K : EGalE K ] = |GalEK|.

Consider that GalEK ≤ GalF K, so by group theory we have [GalF K : GalEK] = |GalF K|/|GalEK|, and so

|GalEK|[GalF K : GalEK] = |GalF K|.

Note that by what we just proved |GalF K| = [K : F ]. Therefore,

|GalEK|[GalF K : GalEK] = |GalF K| = [K : F ] = [K : E][E : F ].

Since [K : E] = |GalEK|, we must have

[E : F ] = [GalF K : GalEK]. Before proving the second part we need a lemma.

48 Lemma 16.8. Let K be a finite-dimensional normal extension field of F and E an intermediate field, which is normal over F , Then there is a surjective homomorphism ∼ of groups θ : GalF K → GalF E whose kernel is GalEK. Moreover, GalF K/GalF E = GalEK. Proof. Define θ : GalF K → GalF E σ 7→ σ|E

We first show that θ is well-defined. We only need to show that σ|E(u) ⊆ E and σ|E is surjective. Note that E is an algebraic extension of F . Let u ∈ E and p(x) be its minimal polynomial over F . Since E is a normal extension of F , so all the roots of p(x) are in E, and since σ(u) is a root of p(x), we conclude that σ(u) ⊆ E. Therefore, σ|E ⊆ E. Moreover, since kerσ|E ⊆ kerσ = {0}, we must have σ(E) ∼= E. So σ(E) is a subspace of E isomorphic to E, and so σ(E) = E. Therefore, σ|E is in GalF E, and thus θ is well- defined. Now, we show that θ is surjective. Note that K is finite dimensional over F and also it is a normal extension of F , so it is splitting field of some polynomial (see Theorem 12.5). Since K is also a splitting field of some polynomial over E, then by Theorem 12.4, τ will be extended to an automorphism σ of K. Therefore, θ(σ) = σ|E = τ, and so τ is in the image of θ. Finally, we show that the kernel of θ is GalEK. If σ ∈ GalEK, then σ|E = id, so σ ∈ kerθ. If σ ∈ kerθ, then σ|E = id, and so σ ∈ GalEK. Therefore, kerθ = GALEK, ∼ and GalF K/GalF E = GalEK.

(2) Assume that GalEK C GalF K, we want to show that E is a normal extension of F . We must show that if p(x) is an irreducible polynomial in F [x] with a root u ∈ E, then all of the roots of p(x) are in E. Since K is a normal extension of F , all of the roots of p(x) are in K. We may assume that p(x) has degree bigger than 1, because otherwise the root of p(x) is only u ∈ E. Let v be a root of p(x) distinct from u. Then by Theorem 15.3, there is a σ ∈ GalF K such that σ(u) = v. Since GalEK C GalF K, for any τ ∈ GalEK, τσ = στ1 for some τ1 ∈ GalEK. Note that

τ(v) = τ(σ(u)) = στ1(u) = σ(u) = v.

Therefore, for any τ ∈ GalEK, we have τ(u) = u, and so u ∈ EGalE K = E. As a result, p(x) splits over E. Converse is just the previous lemma.

Remark. The Galois correspondence ϕ is inclusion-reversing, i.e., if E ⊆ L, then ϕ(L) = GalLK ⊆ GalEK = ϕ(E). √ 3 3 Let K be the splitting ﬁeld√ of x − 2. We want to ﬁnd GalF K. Note that [√Q( 2) : 3 3 3 Q] = 3, and we have that Q( 2) ⊂ K since other roots of x − 2 are not in Q( 2).√ By 3 a theorem, GalF K ⊆ S3. Since |GalF K| < |S3| = 6 and |GalF K| = [K : Q] > [Q( 2) : Q] = 3, we must have [K : F ] = 6 and GalF K = S3.

49 Figure 1:

50 17 Solvability by Radicals

Definition. A field K is said to be a radical extension of a field F if there is a chain of fields F = F0 ⊆ F1 ⊆ F2 ⊆ ... ⊆ Ft = K such that for each i = 1, 2, . . . , t,

Fi = Fi−1(ui) and some powers of ui is in Fi−1.

Deﬁnition. Let f(x) ∈ F [x]. The equation f(x) = 0F is said to be solvable by radicals if there is a radical extension of F that contains a splitting ﬁeld of f(x).

Remark. When we say f(x) = 0 is not solvable by radicals, it means there is no formula (including only ﬁeld operations and extraction of roots) for the solution of f(x) = 0.

17.1 Solvable groups A group is said to be solvable if it has a chain of subgroups

G = G0 ⊆ G1 ⊇ G2 ⊇ · · · ⊇ Gn−1 ⊇ Gn = hei such that each Gi is a normal subgroup of the preceding group Gi−1 and the quotient group Gi−1/Gi is abelian.

Theorem 17.1. 1. For n ≥ 5, the group Sn is not solvable. 2. Every homeomorphic image of a solvable group G is solvable.

Deﬁnition. If f(x) ∈ F [x], then Galois group of the polynomial f(x) is GalF K, where K is a splitting ﬁeld of f(x) over F .

We state Galois Criteria without proof.

Theorem 17.2. (Galois’ Criteria) Let F be a ﬁeld of characteristic 0 and f(x) ∈ F [x]. Then f(x) = 0F is solvable by radicals if and only if the Galois group of f(x) is solvable.

5 Example 17.3. Since S5 is not solvable and Galois group of f(x) = 2x −10x+5 ∈ Q[x] is S5, therefore, f(x) = 0 is not solvable by radicals.

51 18 Roots of Unity

Proposition 18.1. Let K be the splitting ﬁeld of xn − 1. Then the set of all roots of xn − 1 is a multiplicative subgroup of K, moreover, it is cyclic.

Proof. Assume that ζ, τ are roots of xn − 1, then (ζτ)n − 1 = 0, and so the set of all roots of xn − 1 is closed under multiplication. Moreover, if ζ is a root of xn − 1, so is ζ−1, thus the set of all roots of xn − 1 produce a multiplicative subgroup of K. It is known in group theory that any ﬁnite multiplicative subgroup of a ﬁeld is cyclic, and so the set of all roots of xn − 1 is a cyclic group. Any root of xn − 1 is called an nth root of unity. The above proposition states that the set of all nth roots of unity is a cyclic group. Any generator of this cyclic group is called a primitive nth root of unity.

Lemma 18.2. Let F be a ﬁled and ζ a primitive nth root of unity in F . Then F contains a primitive dth root of unity for every positive divisor d of n.

Proof. Let n = dt. Note that (ζt)d = 1, moreover if (ζt)i = (ζt)j, 1 ≤ i < j ≤ d, then (ζt)j−i = 1, and so ζt(j−i) = 1, which is a contradiction since t(j − i) < n. Therefore,

(ζt), (ζt)2,..., (ζt)d−1 are distinct, and so they are the roots of xd − 1, and (ζt) is a primitive dth roots of unity.

Theorem 18.3. Let F be a ﬁeld of characteristic 0 and ζ a primitive nth root of unity in some extension ﬁeld of F . Then K = F (ζ) is a normal extension of F , and GalF K is abelian.

Proof. Consider that all roots of xn − 1 are in K = F (ζ), so K is splitting ﬁeld ﬁeld of n x − 1, therefore by Theorem 12.5, K is a normal extension of F . Let σ, τ ∈ GalF K, then as σ(ζ), τ(ζ) are the roots of xn − 1, and the roots of xn − 1 are the powers of ζ, we conclude that σ(ζ) = ζt and τ(ζ) = ζs. So

σ ◦ (τ(ζ)) = σ(ζs) = ζst and τ ◦ (σ(ζ)) = τ(ζt) = ζts

Therefore, σ ◦ τ = τ ◦ σ.

Theorem 18.4. Let F be a ﬁeld of characteristic 0 that contains a primitive nth root of unity. If u is a root of xn − c ∈ F [x] in some extension ﬁeld of F , then K = F (u) is a normal extension of F , and GalF K is abelian. Proof. If u is a root of xn − c and ζ is a primitive nth roots, then ζiu for every 1 ≤ i ≤ n is a root of xn − c since ((ζiu)n − c = (ζi)nun − c = un − c = 0. Moreover,

1, ζu, ζ2u, . . . , ζn−1u are distinct because if ζiu = ζju, then ζi = ζj, which is contradiction. Therefore, the set of all roots of xn − c is {1, ζu, ζ2u, . . . , ζn−1u}. Consider that F (u) is splitting ﬁeld of xn − c and so F (u) is a normal extension of F . With the same argument as the previous theorem we have that GalF K is abelian.

52 19 Representation Theory

A representation can be thought of as a way to model a group with a concrete group of matrices. After giving the precise deﬁnition, we look at some examples. The general linear group of degree d, GLd(C) is the set of all d × d invertible matrices over C. Deﬁnition. A representation of a group G is a group homomorphism

X : G → GLd(C).

Equivalently, to each g ∈ G is assigned X(g) ∈ GLd(C) such that

1. X(1) = Id the identity matrix in GLd(C), and 2. X(gh) = X(g)X(h) for all g, h ∈ G.

The parameter d is called the degree, or dimension, of the representation.

In the remainder of this course , we only say a matrix representation without men- tioning the group G, if it is clear that we are using G.

Example 19.1. All groups have the trivial representation, which is the one sending every g ∈ G to the matrix (1). This is clearly a representation because X(1) = (1) and X(gh) = (1) = (1)(1) = X(g)X(h) for all g, h ∈ G. We often use the notation 1 to stand for the trivial representation of G.

Example 19.2. Let G = Cn the cyclic group of order n. Let g be a generator for Cn, i.e., 2 n−1 Cn = {1, g, g , . . . , g }.

We aim to ﬁnd all one-dimensional representations of Cn. To identify a group homomorphism form Cn to GLd(C), it is enough to give X(g). Assume that X(g) = (c) be a one-dimensional representation. Then X(1) = X(gn) = X(g)n = (c)n = (cn) = 1. Therefore, c must be a nth root of unity, and it is clear that for every root of unity we have a one-dimensional representation.

Example 19.3. One of the important representation for Sn is deﬁning representation of Sn, which is of degree n. If π ∈ Sn, then we let X(π) = (xi,j)n×n, where ( 1 if π(j) = i xi,j = 0 otherwise.

Let V be a vector space of dimension n over C with a ﬁxed basis {v1, . . . , vn}. Let GL(V ) be the set of all invertible linear transformation of V . Note that for every A =   c1  .  n (ai,j) ∈ GLn(C) and c =  .  ∈ C , we have (Ac)i,1 = ai1c1 + ... + aincn cn

Theorem 19.4. Let V be a vector space of dimension n over C with a ﬁx basis {v1, . . . , vn}. ∼ Then GLn(C) = GL(V ).

53 Proof. Deﬁne α as follows,

α : GLn(C) → GL(V ) A 7→ TA,

where if v = c1v1 + ... + cnvn, then     c1 c1  .   .  TA(c1v1 + ... + cnvn) = (A  . )1,1v1 + ... + (A  . )n,1vn. cn cn

19.1 G-modules and Group algebras Deﬁnition. (1) Let V be a vector space and G be a group. Then V is a G-module if there is a group homomorphism ρ : G → GL(V ). Deﬁnition. (2) The vector space V is a G-module if there is a multiplication, g.v, of elements of V by elements of G such that 1. g.v ∈ V, 2. g.(cv + dw) = c(g.v) + d(g.w), 3. g.(h.v) = (gh).v,

4. e.v = v for all g, h ∈ G, v, w ∈ V , and scalars c, d ∈ C. Why the two deﬁnitions are equivalent? Assume that there is a group homomorphism ρ : G → GL(V ).

Denote the homomorphism ρ(g) by ρg. Then we can deﬁne a multiplication as follows g.v = ρg(v). It is easy to check that this multiplication has all desired properties. Now if we have a vector space with the properties in the second deﬁnition, then ρ : G → GL(V ), g → ρg, where ρg(v) = g.v.

19.2 Action of a group on a set yields a G-module Before we start, let produce a matrix out of a linear transformation form V to V . Let V be a vector space with B = {v1, . . . , vn} as a basis, and let T be a linear transformation from V to V , then [T ]B = [[T (v1)]B ... [T (vn)]B] is an n × n matrix. Moreover, we have ∼ GL(V ) = GLn(C) in which the image of T is [T ]B.

54 Deﬁnition. We say a group G acts on a set S if there is a multiplication

. : G × S → S (g, s) 7→ g.s

such that

1. 1.s = s

2. (gh).s = g.(h.s) for all g, h ∈ G and s ∈ S.

Now assume that G acts on S = {s1, . . . , sn}. Let

CS = C{s1, . . . , sn} = {c1s1 + ... + cnsn : ci ∈ C} consists of all formal linear combination of the elements in S

• CS is a vector space with the following addition and scalar multiplication,

+ : CS × CS → CS (c1s1 + ... + cnsn, d1s1 + ... + dnsn) 7→ (c1 + d1)s1 + ... + (cn + dn)sn

c(c1s1 + ... + cnsn) = (cc1)s1 + ... + (ccn)sn. Note that the set S is a basis for CS as a C-vector space and so the dimension of CS is |S|.

• Now we have that CS is a G-module with the following group homomorphism,

ρ : G → GL(CS) g 7→ ρg,

where ρg(c1s1 + ... + cnsn) = c1(g.s1) + ... + cn(g.sn).  or equivalently we can deﬁne a multiplication as follows,   . : G × CS → CS  (g, c1s1 + ... + cnsn) 7→ c1(g.s1) + ... + cn(g.sn) Moreover, any this G-module produce a representation

ρ X : G → GL(CS) → GLn(C) g 7→ ρg 7→ [ρg]S

Deﬁnition. If G acts on a set S, then the G-module CS as deﬁned above is called permutation representation associated with S.

Example 19.5. Consider that Sn acts on S = {1, 2, . . . , n} as follows,

. : Sn × S → S (σ, i) 7→ σ(i)

55 Therefore, we can consider CS = {c11 + c22 + ... + cnn : ci ∈ C} as a Sn module with the homomorphism ρ : Sn → GL(CS) σ 7→ ρσ, where ρσ(c11 + ... + cnn) = c1(σ.1) + ... + cn(σ.n). And for example, if S = {1, 2, 3}, then  0 1 0  X((1, 2)) = [ρ(1,2)]S =  1 0 0  . 0 0 1

Example 19.6. (Regular representation) Let G = {g1, . . . , gn} be a group, then as G always acts on G as follows, . : G × G → G (g, h) 7→ gh So the corresponding G-module is

CG = {c1g1 + ... + cngn : ci ∈ C} which has the following homomorphism ρ : G → GL(C[G]) g 7→ ρg where ρg(c1g1 + ... + cngn) = c1(gg1) + ... + cn(ggn). 2 3 As an example if G = C4 = {e, g, g , g }, then 2 3 C[C4] = {c1e + c2g + c3g + c4g : ci ∈ C}. Moreover,  0 0 1 0  2  0 0 0 1  2 X(g ) = [ρg ]C4 =    1 0 0 0  0 1 0 0 Example 19.7. (Coset representation of G with respect to H) Let H be a subgroup of G, then G = g1H t ... t gkH, and g1, g2, . . . , gk are called transversal for H. Let

H = {g1H, g2H, . . . , gkH}. Then there is an action of G on H as follows, . : G × H → H (g, giH) 7→ (ggi)H So the corresponding G-module is

CH = {c1(g1H) + ... + ck(gkH): ci ∈ C}. And also for example if H = {e, (2, 3)}, then H = {H, (1, 2)H, (2, 3)H}.  0 1 0  X((1, 2)) = ρ(1,2) =  1 0 0  . 0 0 1

56 20 Reducibility

Deﬁnition. Let V be a G-module. A submodule of V is a subspace W that is closed under the action of G, i.e.,

w ∈ W, g ∈ G ⇒ g.w ∈ W.

In this situation we also say W is G-invariant. (It is equivalent to say if ρ is the group homomorphism of the G-module V , then W is a subspace of V if ρ|W ∈ GL(W )). Example 20.1. Every G-module V has two trivial submodules W = {0} and W = V .

Example 20.2. Consider the C{1,..., n} as a Sn module. Note that

W = C{1 + ... + n} is a subspace of V since for every σ ∈ Sn and w = c(1 + ... + n) ∈ W , we have σ.w = c(σ(1) + ... + σ(n)) ∈ W .

Example 20.3. Let G = {g1, . . . , gn} with group algebra V = C[G]. Let

W = C[g1 + ... + gn].

Note that W is a G-module since for every g ∈ G and c(g1 + ... + gn),

g.(c(g1 + ... + gn)) = c(gg1 + gg2 + ... + ggn) = c(g1 + ... + gn) ∈ W.

Example 20.4. Consider [S ] and W = [P sgn(σ)σ]. Then for every π ∈ S C n C σ∈Sn n and c P sgn(σ)σ, we have σ∈Sn X X X π.c sgn(σ)σ = c( sgn(σ)π ◦ σ) = c( sgn(π−1 ◦ τ)τ) = −1 σ∈Sn σ∈Sn π σ∈Sn X X ±c( sgn(π−1 ◦ τ)τ) = ±c sgn(σ)σ. −1 π σ∈Sn σ∈Sn Negative sign is when sgn(π) = −1.

Deﬁnition. A nonzero G-module V is reducible if it contains a nontrivial submodule W . Otherwise, V is said to be irreducible. Equivalently, V is reducible if it has a basis B in which every g ∈ G is assigned a block matrix of the form

A(g) B(g) X(g) = 0 C(g)

where A(g) are square matrices, all of the same size, and 0 is a nonempty matrix of zeros.

Example 20.5. Let V = C{1, 2, 3} and W = C{1 + 2 + 3}. Note that {1 + 2 + 3} is a basis for W and B = {1 + 2 + 3, 2, 3} is a basis for V . Consider that W is a submodule of V . We want to ﬁnd corresponding representation,

X : G → GL(V ) → GL3(C) g 7→ ρg 7→ [ρg]B

57 Thus, X((1, 2)) = [[ρ(1,2)(1 + 2 + 3)]B [ρ(1,2)(2)]B [ρ(1,2)(3)]B]. We have

ρ(1,2)(1 + 2 + 3) = 1 + 2 + 3,

ρ(1,2)(2) = 1 = 1 + 2 + 3 − 2 − 3

ρ(1,2)(3) = 3 = 3. Therefore,  1 1 0  X((1, 2)) =  0 −1 0  0 −1 1

If you check for any σ ∈ S3, you will see  ∗ ∗ ∗  X(g) =  0 ∗ ∗  0 ∗ ∗

21 Inner product space

Deﬁnition. An inner product on a vector space V is a function

h., .i : V × V −→ C satisfying the following axioms: 1. hu, vi = hv, ui 2. hu + v, wi = hu, wi + hv, wi 3. hcu, vi = chu, vi 4. hu, ui ≥ 0 and hu, ui = 0 if and only if u = 0. A vector space with an inner product is called an inner product space. Moreover, it is clear from the deﬁnition any subspace of an inner product space is an inner product space. Deﬁnition. Two vectors v, w ∈ V are orthogonal if hv, wi = 0.

Deﬁnition. Let y ∈ V where V is an inner product space. Let {v1, . . . , vp} be an orthogonal basis for W . Then the orthogonal projection of y onto a subspace W of V is

hy, v1i hy, v2i hy, vpi projW y = v1 + v2 + ... + vp. hv1, v1i hv2, v2i hvp, vpi

⊥ Note that projW y ∈ W , y − projW y ∈ W . Theorem 21.1. (The Gram-Schmidt process for an inner product space) Given a basis {x1, . . . , xp} for non-zero subspace W of an inner product space V , deﬁne

v1 = x1

hx2,v1i v2 = x2 − v1 hv1,v1i

hx3,v1i hx3,v2i v3 = x3 − v1 − v2 hv1,v1i hv2,v2i

58 . .

hxp,v1i hxp,v2i hxp,vp−1i vp = xp − v1 − v2 − ... − vp−1 hv1,v1i hv2,v2i hvp−1,vp−1i Then {v1, . . . , vp} is an orthogonal basis for W . In addition span{v1, . . . , vk} = span{x1, . . . , xk} for 1 ≤ k ≤ p.

Theorem 21.2. Let W be a subspace of an inner product space V . Then

W ⊥ = {v ∈ V : hv, wi = 0, for all w ∈ W }, the orthogonal complement of W , is also an inner product spaces.

Proof. We show that W ⊥ is a subspace, i.e., if s, z ∈ W ⊥, then cz + s for c ∈ C, is in W ⊥. Note that for any w ∈ W ,

hcz + s, wi = chz, wi + hs, wi = 0, and so cz + s ∈ W .

Deﬁnition. Let V be a vector space with subspaces U and W . Then V is (internal) direct sum of U and W , written V = U ⊕ W if U ∩ W = {0} and every element v ∈ V can be written as v = u + w where u ∈ U and w ∈ W . If V,U and W are G-modules we say U and W are complement of each other.

Theorem 21.3. Let V be an inner product space and W be a subspace of V . Then V = W ⊕ W ⊥.

Proof. If w ∈ W ∩ W ⊥, then for every v ∈ W , we must have hw, vi = 0, so hw, wi = 0, and it implies w = 0.

Moreover, any element v ∈ V , can be written as v = projW v + (v − projW v), where ⊥ projW v ∈ W and (v − projW v) ∈ W .

22 Maschke’s Theorem

Remark. When V = W ⊕ U, then there is a basis for V such that the corresponding homomorphism is of the from

A(g) 0 X(g) = . 0 B(g)

Deﬁnition. If X is a matrix, we say X is the direct sum of A and B, written X = A⊕B, if A 0 . 0 B

Deﬁnition. Let V be a G-module with an orthogonal basis {v1, . . . , vn}. Then if v = c1v1 + ... + cnvn, and w = d1v1 + ... + dnvn, deﬁne

n X hv, wi = cidjδvi,vj , i=1

59 where ( 1 i = j δv ,v = i j 0 otherwise. Now deﬁne, 0 X hv, wi = hgv, gwi. g∈G

Theorem 22.1. Let V be a G-module. Then V is an inner product space with h., .i0 and moreover, if W is a G-submodule of V , then W ⊥ is also a G-module.

Proof. It is easy to check that V is an inner product space with the inner product. We only show that W ⊥ is a G-submodule of V . Let g ∈ G and z ∈ W ⊥, then we should show that h.z ∈ W ⊥, i.e., hhz, wi0 = 0 for every w ∈ W . Note that

0 X hhz, wi = hghz, gwi. g∈G

0 X hz, h−1wi = hgz, gh−1wi. g∈G Let t = gh−1, then g = th. So,

0 X X X 0 0 = hz, h−1wi = hthz, twi = hthz, twi = hghz, gwi = hhz, wi . th∈G t∈G g∈G

Therefore, hz ∈ W ⊥ and so W ⊥ is a G-module.

Theorem 22.2. (Mascheke’s Theorem) Let G be a ﬁnite group and let V be a nonzero G-module. Then V = W (1) ⊕ · · · ⊕ W (k) where each W (i) is an irreducible G-module of V .

Proof. Let dimV = d, we prove the theorem by induction. Let dimV = 1, then V must be an irreducible module. Now assume that the theorem is true for any positive integer less than d and dimV = d > 1. If V is an irreducible module we are done, otherwise it has a nontrivial submodule W and by the previous theorem V = W ⊕ W ⊥. Notice that dimW and dimW ⊥ are less than d, so by induction hypothesis they decompose to irreducible submodules and so does V .

Corollary 22.3. Let G be a group and X be a matrix representation of G of dimension d > 0. Then there is a ﬁxed matrix T such that every matrix X(g), g ∈ G, is of the form

 X(1)(g) 0 ... 0   0 X(2)(g) ... 0  −1   TX(gT ) =  . . .. .   . . . .  0 0 ...X(k)(g)